text:
\section{Introduction}\label{s1}

A map $f:=(f_1,\ldots,f_m):\R^n\to\R^m$ is a {\em polynomial map} if each component $f_i\in\R[{\tt x}} \newcommand{\y}{{\tt y}]:=\R[{\tt x}} \newcommand{\y}{{\tt y}_1,\ldots,{\tt x}} \newcommand{\y}{{\tt y}_n]$. A subset $S$ of $\R^m$ is a {\em polynomial image} of $\R^n$ if there exists a polynomial map $f:\R^n\to\R^m$ such that $S=f(\R^n)$. More generally, the map $f$ is {\em regular} if each component $f_i$ is a regular function of $\R({\tt x}} \newcommand{\y}{{\tt y}):=\R({\tt x}} \newcommand{\y}{{\tt y}_1,\ldots,{\tt x}} \newcommand{\y}{{\tt y}_n)$, that is, $f_i:=\frac{g_i}{h_{i}}$ is a quotient of polynomials such that the zero set of $h_{i}$ is empty. Analogously, a subset $S$ of $\R^m$ is a {\em regular image} of $\R^n$ if it is the image $S=f(\R^n)$ of $\R^n$ given by a regular map $f$.

The present work continues the general study of polynomial and regular images of Euclidean spaces already began in \cite{fg1,fg2}. A celebrated Theorem of Tarski-Seidenberg \cite[1.4]{bcr} says that the image of any polynomial map (and more generally of a regular map) $f:\R^m\rightarrow\R^n$ is a semialgebraic subset $S$ of $\R^n$, that is, it can be written as a finite boolean combination of polynomial equations and inequalities, which we will call a \em semialgebraic \em description. By \em elimination of quantifiers \em $S$ is semialgebraic if it has a description by a first order formula \em possibly with quantifiers\em. Such a freedom gives easy semialgebraic descriptions for topological operations: interiors, closures, borders of semialgebraic sets are again semialgebraic.

In an \em Oberwolfach \em week \cite{g} Gamboa proposed to characterize the semialgebraic sets of $\R^m$ that are polynomial images of $\R^n$ for some $n\geq1$. The open ones deserve a special attention in connection with the real Jacobian Conjecture \cite{j1,j2,p}. The interest of polynomial (and also regular) images arises because there exist certain problems in Real Algebraic Geometry that can be reduced for such sets to the case $S=\R^n$ (see \cite{fu1,fu2}). Examples of such problems are:
\begin{itemize}
\item[$\bullet$]Optimization of polynomial (and/or regular) functions on $S$,
\item[$\bullet$]Characterization of the polynomial (or regular functions) that are positive semidefinite on $S$ (Hilbert's 17th problem and Positivstellensatz),
\item[$\bullet$]Computation of trajectories inside $S$ parametrizable by polynomial (or regular) maps.
\end{itemize}

\subsection{Main result}
We denote the projective space of coordinates $(x_0:x_1:\cdots:x_m)$ with $\R\PP^m$. It contains $\R^m$ as the set of points with $x_0=1$. The hyperplane at infinity $\mathsf{H}_\infty$ has equation $x_0=0$. Given a semialgebraic set $S\subset\R^m$, the \em set of points at infinity of $S$ \em is $S_\infty:=\operatorname{Cl}_{\R\PP^m}(S)\cap\mathsf{H}_\infty$. Our main result in this work is the following.

\begin{thm}\label{main1}
Let $f:\R^n\to\R^m$ be a non-constant polynomial map and denote $S:=f(\R^m)$. Then $S_\infty$ is non-empty and connected.
\end{thm}

It seems a difficult matter to provide a full geometric characterization of all polynomial and/or regular images $S\subset\R^m$. We only know it for the $1$-dimensional case \cite{fe}. Even though, we have approached the representation as polynomial or regular images of ample families of $n$-dimensional semialgebraic sets whose boundaries are piecewise linear. We have focused on: convex polyhedra, their interiors, their exteriors and the closure of their exteriors \cite{fgu3,fu1,fu2,u2}. The proofs are constructive but the arguments are developed \em ad hoc\em. Two main difficulties arise:
\begin{itemize}
\item To develop a strategy to produce an either polynomial or regular map whose image is the desired semialgebraic set.
\item To prove the surjectivity of the constructed map.
\end{itemize}

In \cite{fg1} appear some straightforward properties that a polynomial (resp. regular) image $S\subset\R^m$ must satisfy: $S$ must be pure dimensional, connected, semialgebraic and its Zariski closure must be irreducible. It follows from \cite[3.1]{fg3} that $S$ must be \em irreducible \em in the sense proposed in \cite{fg3}. All these properties follow readily from the fact \cite[3.6]{fgu2}:

\vspace{2mm}
\noindent $(*)$ \em Given two points $p,q\in S$, there exists a polynomial (resp. regular) image $L$ of $\R$ \em (also known as \em parametric semiline\em) \em contained in $S$ and passing through $p,q$\em.

\vspace{2mm}
There are many examples of semialgebraic sets with property $(*)$ that are polynomial images of no $\R^n$. Take $S:=\{0\leq x\leq 1, 0\leq y\}\cup\{0\leq y\leq x\}\subset\R^2$, which satisfies $(*)$. By Theorem \ref{main1} $S$ is a polynomial image of no $\R^n$ because its set of points at infinity is disconnected. Consequently Theorem \ref{main1} provides a new obstruction to be a polynomial image of $\R^n$.

We wondered in \cite[7.3]{fg2} about the number of connected components of the exterior of a polynomial image of dimension $\geq2$. The first author was convinced that the answer was one, but the second author showed in \cite{u1} that this number can be arbitrarily large. Nevertheless, Theorem \ref{main1} is in the vein of our wrong initial position.

\subsection{Strategy of the proof and structure of the article}
The proof of Theorem \ref{main1} involves techniques inspired by those employed by Jelonek in his works \cite{j1,j2} where he studies the geometry of the set of points ${\mathcal S}_f$ at which an either complex or real polynomial map $f:{\mathbb K}} \newcommand{\N}{{\mathbb N}^n\to{\mathbb K}} \newcommand{\N}{{\mathbb N}^m$ is not proper (${\mathbb K}} \newcommand{\N}{{\mathbb N}$ denotes either $\R$ or $\C$). We highlight the following:
\begin{itemize}
\item[$\bullet$]\em Resolution of indeterminacy of rational maps \em defined on projective surfaces.
\item[$\bullet$]Sufficient conditions to guarantee that the intersection of two connected complex projective curves of a complex projective surface is either empty or a singleton.
\item[$\bullet$]A \em `rational' curve selection lemma\em.
\end{itemize}
For the sake of the reader we include a careful exposition of these techniques in Section \ref{s2}. The reader can proceed directly to Section \ref{s3} and refer to the Preliminaries only when needed. In Section \ref{s3} we prove Theorem \ref{main1} in the more general setting of \em quasi-polynomial maps\em. In Section \ref{s4} we show that the set of points at infinity of the image of a general regular map does not need to be connected and we provide some enlightening examples.

\subsection*{Acknowledgements}
The authors thank Prof. Jes\'us M. Ruiz for many helpful discussions and suggestions during the preparation of this article. The authors are also very grateful to S. Schramm for a careful reading of the final version and for the suggestions to refine its redaction.

\section{Preliminaries}\label{s2}

We write ${\mathbb K}} \newcommand{\N}{{\mathbb N}$ to refer indistinctly to $\R$ or $\C$. We denote the hyperplane at infinity of ${\mathbb K}} \newcommand{\N}{{\mathbb N}\PP^m$ with $\mathsf{H}_\infty({\mathbb K}} \newcommand{\N}{{\mathbb N}):=\{x_0=0\}$. Clearly, ${\mathbb K}} \newcommand{\N}{{\mathbb N}\PP^m$ contains ${\mathbb K}} \newcommand{\N}{{\mathbb N}^m$ as the set ${\mathbb K}} \newcommand{\N}{{\mathbb N}\PP^m\setminus\mathsf{H}_\infty({\mathbb K}} \newcommand{\N}{{\mathbb N})=\{x_0=1\}$. If $m=1$, we denote the point at infinity ${\mathbb K}} \newcommand{\N}{{\mathbb N}\PP^1$ with $\{p_\infty\}:=\{x_0=0\}$ and if $m=2$, we write $\ell_\infty({\mathbb K}} \newcommand{\N}{{\mathbb N}):=\{x_0=0\}$ for the line at infinity of ${\mathbb K}} \newcommand{\N}{{\mathbb N}\PP^2$. We use freely that the real projective space $\R\PP^m$ can be immersed in $\R^k$ for $k$ large enough as an affine non-singular real algebraic variety \cite[3.4.4]{bcr}. Thus, the closure in $\R\PP^m$ of a semialgebraic subset of $\R^m$ is again a semialgebraic set. It will be useful to understand real algebraic objects as fixed parts under conjugation of complex algebraic objects that are invariant under conjugation.

\subsection{Invariant projective objects.}\label{inv}
For each $n\geq1$ denote the complex conjugation with
$$
\sigma:=\sigma_n:\C\PP^n\to\C\PP^n,\ z=(z_0:z_1:\cdots:z_n)\mapsto\overline{z}=(\overline{z_0}:\overline{z_1}:\cdots:\overline{z_n}).
$$
Clearly, $\R\PP^n$ is the set of fixed points of $\sigma$. A set $A\subset\C\PP^n$ is called \em invariant \em if $\sigma(A)=A$. If $Z\subset\C\PP^n$ is an invariant non-singular (complex) projective variety, then $Z\cap\R\PP^n$ is a non-singular (real) projective variety. We say that a rational map $h:\C\PP^n\dashrightarrow\C\PP^m$ is \em invariant \em if $h\circ\sigma_n=\sigma_m\circ h$. Of course, $h$ is invariant if its components are homogeneous polynomials with real coefficients, so it provides by restriction a real rational map $h|_{\R\PP^n}:\R\PP^n\dashrightarrow\R\PP^m$. We use freely usual concepts of Algebraic Geometry such as: rational map, regular map, divisor, blow-up, etc. and refer the reader to \cite{ha,m,sh1,sh2} for further details. For the sake of the reader we denote complex dimension with $\dim_\C(\cdot)$ and real dimension with $\dim_\R(\cdot)$. Recall the following fact concerning the regularity of rational maps defined on a non-singular projective curve \cite[7.1]{m}.

\begin{lem}\label{curve}
Let $Z\subset\C\PP^n$ be a non-singular projective curve and $F:Z\dashrightarrow\C\PP^m$ a rational map. Then $F$ extends to a regular map $F':Z\to\C\PP^m$. In addition, if $Z,F$ are invariant, so is $F'$.
\end{lem}

One of the main tools is the resolution of indeterminacy of an invariant rational map. We provide a careful presentation of this well-known tool taking care of invariance.

\subsection{Resolution of indeterminacy of an invariant rational map.}\label{res}
Let $Z_0\subset\C\PP^n$ be an invariant non-singular projective variety of dimension $d$ and
$$
F_\C:=(F_1:\cdots:F_m):Z_0\dashrightarrow\C\PP^m
$$
an invariant rational map. To compute the \em set of indeterminacy \em of $F_\C$ one proceeds as follows \cite[III.1.4]{sh1}. Consider for each $i=0,\ldots,m$ the divisor $D_i$ in $Z_0$ defined by $F_i$ and let $|D|:=\operatorname{hcd}\{D_0,D_1,\ldots,D_m\}$ be the highest common divisor of the divisors $D_0,D_1,\ldots,D_m$. The divisors $D_i':=D_i-|D|$ are relatively prime. By \cite[III.1.4.Thm.2]{sh1} the map $F_\C$ fails to be regular exactly at the points of the invariant set $Y_\C:=\bigcap_{i=0}^m\operatorname{supp}(D_i')$, which has dimension $\leq d-2$. As $F_\C$ is invariant, it can be restricted to a real rational map $F_\R:Z_0\cap\R\PP^n\dashrightarrow\R\PP^m$ whose set of indeterminacy is $Y_\R:=Y_\C\cap\R\PP^2$.

We assume that $Z_0$ has dimension $2$. As it is well-known, $F_\C:Z_0\dashrightarrow\C\PP^m$ admits an invariant resolution. Namely,

\paragraph{}\label{clue} There exist:
\em\begin{itemize}
\item[(i)] An invariant non-singular projective surface $Z_1\subset\C\PP^k$ for some $k\geq2$.
\item[(ii)] An invariant (composition of a) sequence of blow-ups $\pi_\C:Z_1\to Z_0\subset\C\PP^n$ such that $\pi_\C|_{Z_1\setminus\pi_\C^{-1}(Y_\C)}:Z_1\setminus\pi_\C^{-1}(Y_\C)\to Z_0\setminus Y_\C$ is a biregular isomorphism and
$$
Y_\C=\{p\in Z_0:\ \#\pi_\C^{-1}(p)>1\}.
$$
\item[(iii)] An invariant regular map $\widehat{F}_\C:Z_1\to\C\PP^m$ such that
$$
\widehat{F}_\C|_{Z_1\setminus\pi_\C^{-1}(Y_\C)}=F_\C\circ\pi_\C|_{Z_1\setminus\pi_\C^{-1}(Y_\C)}.
$$
\end{itemize}

In addition, for each $y\in Y_\C$ the irreducible components of $\pi_\C^{-1}(y)$ are non-singular projective curves $K_{i,y}$ that are biregularly equivalent to $\C\PP^1$ (via regular maps $\Phi_{i,y}:\C\PP^1\to K_{i,y}$ that are invariant for invariant $K_{i,y}$) and satisfy:
\begin{itemize}
\item[(iv)] If $y\in Y_\C\setminus Y_\R$, then $\sigma(K_{i,y})=K_{i,\sigma(y)}$ and $K_{i,y}\cap\R\PP^k=\varnothing$,
\item[(v)] If $y\in Y_\R$, then either $K_{i,y}\cap\R\PP^k=\varnothing$ and there exists $j\neq i$ such that $\sigma(K_{i,y})=K_{j,y}$ or $\sigma(K_{i,y})=K_{i,y}$ and $C_{i,y}=K_{i,y}\cap\R\PP^k$ is a non-singular projective curve biregularly equivalent to $\R\PP^1$ (via $\phi_{i,y}:=\Phi_{i,y}|_{\R\PP^1}:\R\PP^1\to C_{i,y}$).
\end{itemize}
\em A triple $(Z_1,\pi_\C,\widehat{F}_\C)$ satisfying the previous properties is an \em invariant resolution for $F_\C$\em.

Let us recall some terminology and results concerning blow-ups of non-singular projective varieties at non-singular centers, from which \ref{clue} follows readily.

\subsection{Blow-up with a non-singular variety as center}\label{blu}\setcounter{paragraph}{0}

Let $Z_0\subset\C\PP^n$ be a non-singular irreducible projective variety and $Y\subset Z_0$ a non-singular subvariety. Let $H_1,\ldots,H_m$ be a system of homogeneous polynomials of the same degree that generates an ideal $I$ whose saturation
$$
\overline{I}:=\{H\in\C[{\tt z}} \renewcommand{\t}{{\tt t}]:=\C[{\tt z}} \renewcommand{\t}{{\tt t}_0,{\tt z}} \renewcommand{\t}{{\tt t}_1,\ldots,{\tt z}} \renewcommand{\t}{{\tt t}_n]:\ ({\tt z}} \renewcommand{\t}{{\tt t})^kH\subset I\quad\text{for some $k\geq0$}\}
$$
equals the ideal ${\mathcal J}(Y)$ of (homogeneous) polynomials of $\C[{\tt z}} \renewcommand{\t}{{\tt t}]$ vanishing identically on $Y$.

\paragraph{}\label{propblu} The \em blow-up $\operatorname{Bl}_{Y}(Z_0)$ of $Z_0$ with center $Y$ \em is the closure in $Z_0\times\C\PP^{m-1}$ of the set
$$
\{(z;(H_1(z):\cdots:H_m(z)))\in (Z_0\setminus Y)\times\C\PP^{m-1}\}
$$
together with the projection $\pi:\operatorname{Bl}_{Y}(Z_0)\subset Z_0\times\C\PP^{m-1}\to Z_0,\ (z;y)\mapsto z$ (see \cite[7.18]{ha} and \cite[VI.2.2]{sh2}). Recall the following facts:
\begin{itemize}
\item[$\bullet$]$\operatorname{Bl}_{Y}(Z_0)$ is a non-singular irreducible projective variety of the same dimension as $Z_0$, independent of the choices made in the process.
\item[$\bullet$]$\pi^{-1}(Y)$ is a non-singular hypersurface of $\operatorname{Bl}_{Y}(Z_0)$.
\item[$\bullet$]$\pi|_{\operatorname{Bl}_{Y}(Z_0)\setminus\pi^{-1}(Y)}:\operatorname{Bl}_{Y}(Z_0)\setminus\pi^{-1}(Y)\to Z_0\setminus Y$ is a biregular isomorphism.
\item[$\bullet$]$Y$ admits a finite cover by affine open subsets $\{U_\alpha\}_\alpha$ satisfying $\pi^{-1}(U_\alpha)\cong U_\alpha\times\C\PP^{r-1}$ where $r:=\dim_\C(Z_0)-\dim_\C(Y)$. In particular, the fiber of each $y\in Y$ is a projective space $\C\PP^{r-1}$.
\item[$\bullet$]If $Y_1,\ldots,Y_r$ are the irreducible components of $Y$, then they are pairwise disjoint and non-singular and it holds
$$
\operatorname{Bl}_Y(Z_0)\cong\operatorname{Bl}_{Y_r}(\cdots\operatorname{Bl}_{Y_2}(\operatorname{Bl}_{Y_1}(Z_0))\cdots).
$$
\end{itemize}
If $Z_0,Y$ are invariant, $\operatorname{Bl}_{Y}(Z_0)$ can be assumed invariant, too, by choosing an ideal $I$ with saturation ${\mathcal J}(Y)$ and whose generators are invariant (given any family of generators, consider the real and the imaginary parts of all of them). If we consider the immersion of $\operatorname{Bl}_{Y}(Z_0)$ in some $\C\PP^N$ using Segre's map, also the regular map $\pi:\operatorname{Bl}_{Y}(Z_0)\to Z_0$ is invariant.

\paragraph{}\label{curve2} Assume that $Z_0$ is an invariant non-singular projective surface and $Y$ is a finite invariant subset. Consider the invariant blow-up $(\operatorname{Bl}_{Y}(Z_0),\pi)$ of $Z_0$ with center $Y$:

\indent (i) For each $y\in Y$ the fiber $\pi^{-1}(y)$ is a $\C\PP^1$. If $y\in Y\cap\R\PP^d$, there exists an invariant biregular equivalence between $\pi^{-1}(y)$ and $\C\PP^1$. If $y\in Y\setminus\R\PP^d$, then $\pi^{-1}(y)\cap\R\PP^N=\varnothing$.

\indent (ii) If $C\subset Z_0$ is a non-singular curve not contained in the center $Y$, its strict transform
$$
\widetilde{C}:=\operatorname{Cl}_{\operatorname{Bl}_{Y}(Z_0)}(\pi^{-1}(C\setminus Y))\subset\pi^{-1}(C)
$$
is as well a non-singular curve. In addition, if there exists an invariant biregular equivalence $\Phi:\C\PP^1\to C$, the strict transform $\widetilde{C}\subset\pi^{-1}(C)$ of $C$ under $\pi$ is invariant and there exists an invariant biregular equivalence $\Psi:\C\PP^1\to\widetilde{C}$.

\begin{proof}[Sketch of proof of statement \em \ref{clue}]
To solve the indeterminacy of the rational map $F_\C:Z_0\dashrightarrow\C\PP^m$, one blows the set of points of indeterminacy $Y_\C$ of $F_\C$ up and considers the composition $G$ of $F_\C$ with the previous sequence of blowing-ups next \cite[IV.3.3.Thm.3]{sh1}. If $G$ is regular, the process is concluded. Otherwise one applies the previous procedure to $G$. In finitely many steps one achieves a regular map and the process finishes. By \ref{blu} we may assume that each rational map involved in the process is invariant, so in the last step of the process we obtain:
\begin{itemize}
\item An invariant non-singular projective surface $Z_1\subset\C\PP^k$ for some $k\geq2$,
\item An invariant sequence of blow-ups $\pi_\C:Z_1\to Z_0\subset\C\PP^n$ and
\item An invariant regular map $\widehat{F}_\C:Z_1\to\C\PP^m$ such that
$$
\widehat{F}_\C|_{Z_1\setminus\pi_\C^{-1}(Y_\C)}=F_\C\circ\pi_\C|_{Z_1\setminus\pi_\C^{-1}(Y_\C)}.
$$
\end{itemize}
The triple $(Z_1,\pi_\C,\widehat{F}_\C)$ satisfies conditions (i) to (iii). The fiber of each point of $Y_\C$ under $\pi$ is a complex projective curve by \ref{blu} whose irreducible components are non-singular rational curves while the fiber of each point of $Z_0\setminus Y_\C$ under $\pi$ is a singleton. Thus, $Y_\C$ is the fundamental set of $\pi_\C$, so (ii) holds. Assertions (iv) and (v) are straightforward consequences of \ref{blu}.
\end{proof}

\subsection{Projective curves intersecting each other in a singleton.}\setcounter{paragraph}{0}
It will be useful to know sufficient conditions that guarantee that the intersection of two connected complex projective curves of a complex projective surface is either empty or a singleton. The proof of the following result is deeply inspired by the proofs of \cite[4.6]{j1} and \cite[3.1]{j2}.

\begin{lem}\label{crack}
Let $X$ be a complex projective surface. Assume that:
\begin{itemize}
\item[$\bullet$]$U\subset X$ is a connected orientable manifold such that
$$
H_1(U;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=H_2(U;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0,
$$
\item[$\bullet$]$U$ is dense in $X$ and the complement $A:=X\setminus U$ is a complex projective curve.
\end{itemize}
Let $C_1,C_2\subset A$ be two connected, complex projective curves without common irreducible components. Then the intersection $C_1\cap C_2$ is either the empty set or a singleton.
\end{lem}
\begin{proof}
The proof is conducted in several steps:

\paragraph{} We prove first: $H^{1}(A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$.

Assume that $X$ is a compact polyhedron of dimension $4$ and $A$ a closed subpolyhedron of $X$. As $U$ is an orientable real manifold of dimension $4$, by Lefschetz duality \cite[6.1.11 \& 6.2.19]{sp} we have $H^{i}(X,A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong H_{4-i}(U;{\mathbb Z}} \newcommand{\R}{{\mathbb R})$ for $i=0,\ldots,4$. By the long exact sequence of cohomology \cite[5.4.13]{sp}
$$
H^1(X,A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to H^1(X;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to H^{1}(A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to H^2(X,A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong H_2(U;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0,
$$
so $H^1(X;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to H^{1}(A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})$ is an epimorphism. As $U$ is a connected open dense subset of $X$, there exists by \cite[4.7]{j1} an epimorphism $0=H_1(U;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to H_1(X,{\mathbb Z}} \newcommand{\R}{{\mathbb R})$, so $H_1(X,{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$. By the universal-coefficient theorem for cohomology \cite[5.5.3]{sp}
$$
H^1(X,{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong{\rm Hom}(H_1(X,{\mathbb Z}} \newcommand{\R}{{\mathbb R}),{\mathbb Z}} \newcommand{\R}{{\mathbb R})\oplus{\rm Ext}({\mathbb Z}} \newcommand{\R}{{\mathbb R},{\mathbb Z}} \newcommand{\R}{{\mathbb R}),
$$
so by \cite[5.5.1]{sp} $H^{1}(X;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$. Consequently, $H^{1}(A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$.

\paragraph{}\label{retract}\em Let $C\subset A$ be a projective algebraic curve. Then $H^1(C;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$\em. In particular, it holds $H^1(C_1\cup C_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$.

Let $C'$ be the union of the irreducible components of $A$ not contained in $C$. Clearly, ${\mathcal F}:=C\cap C'$ is a finite set. As $C$ and $C'$ are analytic sets, they are locally contractible, so for each $x\in{\mathcal F}$ there exists a neighborhood $V^x$ in $A$ such that:
\begin{itemize}
\item[$\bullet$]$V^x\cap C$ and $V^x\cap C'$ have the singleton $\{x\}$ as a deformation retract,
\item[$\bullet$]$V^x\cap C\cap C'=\{x\}$ and
\item[$\bullet$]$V^{x_1}\cap V^{x_2}=\varnothing$ if $x_1,x_2\in{\mathcal F}$ and $x_1\neq x_2$.
\end{itemize}
It holds that $V:=C\cup\bigcup_{x\in{\mathcal F}}(V^x\cap C')$ and $W:=C'\cup\bigcup_{x\in{\mathcal F}}(V^x \cap C)$ are open subsets of $A$ such that $V\cup W=A$, $V\cap W=\bigcup_{x\in{\mathcal F}}(V^x \cap (C\cup C'))$ and $C,C'$ are respective deformation retracts of $V$ and $W$. In addition, ${\mathcal F}$ is a deformation retract of $V\cap W$. By Mayer-Vietoris' exact sequence for cohomology \cite[5.4.9]{sp}
\begin{multline*}
0=H^1(A;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=H^1(V\cup W;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to H^1(V;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\oplus H^1(W;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\\
\to H^1(V\cap W;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong H^1({\mathcal F};{\mathbb Z}} \newcommand{\R}{{\mathbb R}).
\end{multline*}
As ${\mathcal F}$ is a finite set, $H^1({\mathcal F};{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$, so $H^1(C;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong H^1(V;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$.

\paragraph{}\em$C_1\cap C_2$ is either empty or a singleton\em.

Assume $C_1\cap C_2\neq\varnothing$. Let $V_1,V_2$ be two open subsets of $C_1\cup C_2$ such that
\begin{itemize}
\item[$\bullet$]$C_i\subset V_i$ is a deformation retract of $V_i$ for $i=1,2$,
\item[$\bullet$]$V_1\cup V_2=C_1\cup C_2$ and $C_1\cap C_2$ is a deformation retract of $V_1\cap V_2$.
\end{itemize}
(for the construction of $V_1,V_2$ proceed similarly to \ref{retract}). By Mayer-Vietoris' exact sequence for reduced cohomology \cite[5.4.8 \& p.240]{sp} applied to the open subsets $V_1$ and $V_2$ of $C_1\cup C_2$ (whose intersection $V_1\cap V_2\supset C_1\cap C_2\neq\varnothing$), we deduce
\begin{multline*}
\tilde{H}^{0}(C_1;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\oplus\tilde{H}^{0}(C_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong\tilde{H}^{0}(V_1;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\oplus\tilde{H}^{0}(V_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\\
\to\tilde{H}^{0}(V_1\cap V_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong\tilde{H}^{0}(C_1\cap C_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to\tilde{H}^1(V_1\cup V_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\\
\cong\tilde{H}^1(C_1\cup C_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\cong H^1(C_1\cup C_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0.
\end{multline*}
As $C_1,C_2$ are connected, $\tilde{H}^{0}(C_i;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$, so $\tilde{H}^{0}(C_1\cap C_2;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$. Thus, the finite set $C_1\cap C_2$ is connected, so it is a singleton.
\end{proof}
\begin{sexample}\label{bouquet}\em
Let ${\mathcal F}\subset\C^2$ be a finite set and $U:=\C^2\setminus{\mathcal F}$ its complement. Then $H_1(U,{\mathbb Z}} \newcommand{\R}{{\mathbb R})=H_2(U,{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$.

By Hurewicz's theorem $H_1(U,{\mathbb Z}} \newcommand{\R}{{\mathbb R})$ is the abelianization of $\pi_1(U)=0$, so $H_1(U,{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$. We identify $\C^2\equiv\R^4$. To compute $H_2(U,{\mathbb Z}} \newcommand{\R}{{\mathbb R})$, we may assume ${\mathcal F}:=\{p_1,\ldots,p_r\}$ where $p_k:=(2k-1,0,0,0)$. Notice that $D_r:=\bigcup_{i=1}^r{\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_i}$ where ${\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_i}:=\{x\in\R^4:\ \|x-p_i\|=1\}$ is a deformation retract of $U\equiv\R^4\setminus{\mathcal F}$. Observe
$$
{\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_i}\cap{\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_j}=\begin{cases}
\{p_{ij}:=\frac{p_i+p_j}{2}\}&\text{if $|i-j|=1$},\\
\varnothing&\text{if $|i-j|>1$}.
\end{cases}
$$
Denote $D_k:=\bigcup_{i=1}^k{\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_i}$ and observe $H_2(D_1;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=H_2({\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_1};{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$. By induction hypothesis, we assume $H_2(D_{r-1};{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$. By Mayer-Vietoris' exact sequence for homology \cite[\S4.6]{sp}
\begin{multline*}
0=H_2(D_{r-1},{\mathbb Z}} \newcommand{\R}{{\mathbb R})\oplus H_2({\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_r};{\mathbb Z}} \newcommand{\R}{{\mathbb R})\to H_2(D_r;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=H_2(U;{\mathbb Z}} \newcommand{\R}{{\mathbb R})\\
\to H_1(D_{r-1}\cap{\mathbb S}} \newcommand{\D}{{\mathbb D}^3_{p_r},{\mathbb Z}} \newcommand{\R}{{\mathbb R})=H_1(\{p_{r-1,r}\};{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0,
\end{multline*}
so $H_2(U;{\mathbb Z}} \newcommand{\R}{{\mathbb R})=0$, as required.\qed
\end{sexample}

\begin{sremark}\label{crackc}\em
Lemma \ref{crack} applies if $U$ is homeomorphic to the complement in $\C^2$ of a finite subset.
\end{sremark}

\subsection{Rational curve selection lemma.}\setcounter{paragraph}{0}
To finish this section we present the following variation of the curve selection lemma adapted to the situations we will approach later. We refer the reader to \cite[4.7]{jk} for a result of similar nature.

\begin{lem}\label{polcurve}
Let $f:\R^n\to\R^m$ be a regular map and $S:=f(\R^n)$. Let $S'\subset S$ be a semialgebraic dense subset of $S$ and $p\in\operatorname{Cl}_{\R\PP^m}(S)\setminus S$. Then there exist (after reordering the variables of $\R^n$) a rational path $\alpha:=(\pm\t^{k_1},\t^{k_2}{\tt p}_2,\ldots,\t^{k_n}{\tt p}_n)\in\R(\t)^n$ where
\begin{itemize}
\item[$\bullet$]$k_i\in{\mathbb Z}} \newcommand{\R}{{\mathbb R}$, $k_1=\min\{k_1,\ldots,k_n\}<0$,
\item[$\bullet$]${\tt p}_i\in\R[\t]$ and ${\tt p}_i(0)\neq 0$ for $i=2,\ldots,n$
\end{itemize}
and an integer $r\geq1$ such that for each $\beta\in(\t)^r\R[\t]^n$
\begin{itemize}
\item[(i)] $p=\lim_{t\to0^+}(f\circ(\alpha+\beta))(t)$ and
\item[(ii)] $(f\circ(\alpha+\beta))(t)\subset S'$ for $t>0$ small enough.
\end{itemize}
\end{lem}

Before proving the previous result, we need a technical lemma.
\begin{lem}\label{order}
Let $F\in\R[{\tt x}} \newcommand{\y}{{\tt y}]$ be a polynomial that is not identically zero and let $g\in\R((\t))^n$. Then for each $s\geq1$ there exists $r\geq1$ such that if $h\in(\t)^r\R[[\t]]^n$, we have $F(g)-F(g+h)\in(\t)^s\R[[\t]]$.
\end{lem}
\begin{proof}
Write $g:=\frac{g'}{\t^k}$ where $k\geq 0$ and $g'\in\R[[\t]]^n$. Let ${\tt z}} \renewcommand{\t}{{\tt t}$ and $\y:=(\y_1,\ldots,\y_n)$ be variables. Write $F({\tt x}} \newcommand{\y}{{\tt y}+{\tt z}} \renewcommand{\t}{{\tt t}\y)=F({\tt x}} \newcommand{\y}{{\tt y})+{\tt z}} \renewcommand{\t}{{\tt t} H({\tt x}} \newcommand{\y}{{\tt y},\y,{\tt z}} \renewcommand{\t}{{\tt t})$ where $H\in\R[{\tt x}} \newcommand{\y}{{\tt y},\y,{\tt z}} \renewcommand{\t}{{\tt t}]$ is a polynomial of degree $d$. Let $r:=s+kd$ and observe that if $h\in(\t)^r\R[[\t]]^n$, we may write $h:=\t^rh'$ where $h'\in\R[[\t]]$ and
$$
F(g+h)-F(g)=\t^rH\Big(\frac{g'}{\t^k},h',\t^r\Big).
$$
Observe that the order of the series $F(g+h)-F(g)$ is $\geq r-kd=s$, as required.
\end{proof}

\begin{proof}[Proof of Lemma \em\ref{polcurve}]
The proof is conducted in several steps:

\paragraph{} We may assume: \em $S'$ is open in $S$\em.

As $S'$ is dense in $S$ and $S$ is pure dimensional because it is the image of $\R^n$ under a regular map, it holds that $\operatorname{Cl}_{\R^{m}}(S\setminus S')$ has dimension $\leq\dim_\R(S)-1$. Thus, $S'':=S\setminus\operatorname{Cl}_{\R^m}(S\setminus S')\subset S'$ is dense and open in $S$. Changing $S'$ with $S''$, we may assume that $S'$ is open in $S$.

\paragraph{} \em There exists a Nash path $\lambda:(-1,1)\to\R\PP^n$ such that $(f\circ\lambda)(0,1)\subset S'$, $\lim_{t\to0}\lambda(t)\!=q\in\R\PP^n\setminus\R^n$ and $\lim_{t\to0}(f\circ\lambda)(t)=p$\em.

As $p\in\operatorname{Cl}_{\R\PP^m}(S')\setminus S'$, there exists by the Nash curve selection lemma \cite[8.1.13]{bcr} a Nash path $\gamma:(-1,1)\to\R\PP^m$ such that $\gamma((0,1))\subset S'$ and $\gamma(0)=p$. Let $\{x_k\}_k\subset\R^n$ and $\{t_k\}_{k\geq1}\subset(0,1)$ be sequences such that $\lim_{k\to\infty}t_k=0$ and $f(x_k)=\gamma(t_k)$ for all $k\geq1$. We may assume that $\{x_k\}_{k\geq1}$ converges to $q\in\R\PP^n$. As $S=f(\R^n)$ and $p=\gamma(0)\in\operatorname{Cl}_{\R\PP^m}(S)\setminus S$, we have $q\in\R\PP^n\setminus\R^n=\mathsf{H}_\infty(\R)$.

As $\dim_\R(f^{-1}(\gamma((0,1))))\geq\dim_\R(\gamma((0,1)))=1$ and
$$
q\in\operatorname{Cl}_{\R\PP^n}(f^{-1}(\gamma((0,1))))\setminus(f^{-1}(\gamma((0,1)))),
$$
there exists a Nash path $\lambda:(-1,1)\to\R\PP^n$ such that $\lambda((0,1))\subset f^{-1}(\gamma((0,1)))$ and $\lambda(0)=q$.

\paragraph{} \em Construction of the integer $k_1$\em. After reparametrizing $\gamma$, we may assume $f\circ\lambda=\gamma$, so there exist
$$
\lambda_0,\lambda_1,\ldots,\lambda_n,\gamma_0,\gamma_1,\ldots,\gamma_m\in\R[[\t]]
$$
such that
$$
f\circ\lambda=f\Big(\frac{\lambda_1}{\lambda_0},\ldots,\frac{\lambda_n}{\lambda_0}\Big)=\Big(\frac{\gamma_1}{\gamma_0},\ldots,\frac{\gamma_m}{\gamma_0}\Big)=\gamma,
$$
$\lim_{t\to0^+}(1:\lambda(t))=q$ and $\lim_{t\to0^+}(1:\gamma(t))=p$. As $q\in\mathsf{H}_\infty(\R)$, we may assume after reordering the variables $x_1,\ldots,x_n$ that the order
$$
k_1:=\omega\Big(\frac{\lambda_1}{\lambda_0}\Big)=\min\Big\{\omega\Big(\frac{\lambda_i}{\lambda_0}\Big):\ i=1,\ldots,n\Big\}<0.
$$
After a change of the type $\t\mapsto\t u(\t)$ where $u\in\R[[\t]]$ is a unit, we assume $\frac{\lambda_1}{\lambda_0}=\pm\t^{k_1}$ with $k_1<0$. Write
$$
\frac{\lambda_i}{\lambda_0}=t^{k_i'}\rho_i\quad\text{where }\begin{cases}
\text{$k_i'\in{\mathbb Z}} \newcommand{\R}{{\mathbb R}$, $\rho_i\in\R[[\t]]$ and $\rho_i(0)\neq 0$}&\text{if $\frac{\lambda_i}{\lambda_0}\neq0$,}\\
\text{$k_i'=0$ and $\rho_i=0$}&\text{if $\frac{\lambda_i}{\lambda_0}=0$.}
\end{cases}
$$

\paragraph{}\em Construction of the integer $r$\em. Write $f:=(\frac{f_1}{f_0},\ldots,\frac{f_m}{f_0})$ where $f_i\in\R[{\tt x}} \newcommand{\y}{{\tt y}]$ and $f_0$ does not vanish on $\R^n$. By Lemma \ref{order} there exists $r_0\geq 1$ such that if $\mu:=(\mu_1,\ldots,\mu_n)\in(\t)^{r_0}\R[[\t]]^n$, then
\begin{multline*}
\lim_{t\to0^+}(1:\lambda(t)+\mu(t))=q\\
\text{and}\quad\lim_{t\to0^+}\Big(1:\frac{f_1}{f_0}(\lambda(t)+\mu(t)):\cdots:\frac{f_m}{f_0}(\lambda(t)+\mu(t))\Big)=p.
\end{multline*}
Since $S'$ is open in $S$, also $f^{-1}(S')$ is open in $\R^n$. As
$$
\lambda((0,1))\subset f^{-1}(\gamma((0,1)))\subset f^{-1}(S'),
$$
there exist finitely many polynomials $g_1,\ldots,g_q\in\R[{\tt x}} \newcommand{\y}{{\tt y}]$ such that
$$
\lambda((0,\varepsilon_1))\subset\{g_1>0,\ldots,g_q>0\}\subset f^{-1}(S')
$$
for some $\varepsilon_1>0$ small enough. Thus, $g_i\circ\lambda=a_it^{\ell_i}+\cdots$ where $a_i>0$ and $\ell_i\in{\mathbb Z}} \newcommand{\R}{{\mathbb R}$. By Lemma \ref{order} there exists $r\geq\max\{r_0,k_2',\ldots,k_n'\}+1$ such that if $\eta\in(\t)^r\R[[\t]]$, then
$$
(g_i\circ(\lambda+\eta))-(g_i\circ\lambda)\in(\t)^s\R[[\t]]
$$
where $s:=\max\{0,\ell_1,\ldots,\ell_q\}+1$. Consequently, if $\eta\in(\t)^r\R[[\t]]$, each series $g_i\circ(\lambda+\eta)>0$ for $t>0$ small enough, so $(\lambda+\eta)(t)\in f^{-1}(S)$ for $t>0$ small enough.

\paragraph{}\em Construction of the rational path $\alpha$\em. Choose $\mu:=(0,\mu_2,\ldots,\mu_n)\in(\t)^r\R[[\t]]^n$ such that $\t^{-k_i'}\,(\frac{\lambda_i}{\lambda_0}+\mu_i)=\rho_i+\t^{-k_i'}\mu_i\in\R[\t]\setminus\{0\}$. Write $\t^{k_i'}(\rho_i+\t^{-k_i'}\mu_i)=\t^{k_i}{\tt p}_i$ where $k_i\in{\mathbb Z}} \newcommand{\R}{{\mathbb R}$, ${\tt p}_i\in\R[\t]$ and ${\tt p}_i(0)\neq0$. The rational path $\alpha:=(\pm\t^{k_1},\t^{k_2}{\tt p}_2,\ldots,\t^{k_n}{\tt p}_n)\in\R(\t)^n$ and the integer $r$ satisfy the conditions in the statement.
\end{proof}

\section{Connectedness of the set of points at infinity of a polynomial image}\label{s3}

\subsection{Proof of Theorem \ref{main1}}\label{s3a}\setcounter{paragraph}{0}

The purpose of this section is to prove Theorem \ref{main1}. We approach this result in the more general framework of \em quasi-polynomial maps\em. Given a regular map
$$
f:=\Big(\frac{f_1}{f_0},\cdots,\frac{f_m}{f_0}\Big):\R^n\to\R^m
$$
where each $f_i\in\R[{\tt x}} \newcommand{\y}{{\tt y}]$, consider the invariant rational map
$$
F_\C:\C\PP^n\dashrightarrow\C\PP^m,\ x:=(x_0:x_1:\cdots:x_n)\mapsto(F_0(x):F_1(x):\cdots:F_m(x))
$$
where $F_i({\tt x}} \newcommand{\y}{{\tt y}_0:{\tt x}} \newcommand{\y}{{\tt y}_1:\cdots:{\tt x}} \newcommand{\y}{{\tt y}_n):={\tt x}} \newcommand{\y}{{\tt y}_0^df_i(\frac{{\tt x}} \newcommand{\y}{{\tt y}_1}{{\tt x}} \newcommand{\y}{{\tt y}_0},\ldots,\frac{{\tt x}} \newcommand{\y}{{\tt y}_n}{{\tt x}} \newcommand{\y}{{\tt y}_0})$ and $d:=\max_{i=0,\ldots,m}\{\deg(f_i)\}$. Let $Y_\C$ be the set of indeterminacy of $F_\C$ and write $F_0={\tt x}} \newcommand{\y}{{\tt y}_0^eF_0'$ where $e\geq0$ and $F_0'\in\R[{\tt x}} \newcommand{\y}{{\tt y}_0,{\tt x}} \newcommand{\y}{{\tt y}_1,\ldots,{\tt x}} \newcommand{\y}{{\tt y}_n]$ is a homogeneous polynomial that is not divisible by ${\tt x}} \newcommand{\y}{{\tt y}_0$. Observe that $F_\C$ can be restricted to a rational map $F_\R:\R\PP^n\dashrightarrow\R\PP^m$ whose set of indeterminacy is $Y_\R:=Y_\C\cap\R\PP^m$.

\begin{defn}\label{qp}\em
We say that $f$ is a \em quasi-polynomial map \em if $F_\C(\mathsf{H}^n_\infty(\C))\subset\mathsf{H}_\infty^m(\C)$ and no real indeterminacy point of $F_\C$ belongs to $\{F_0'=0\}$, that is, $Y_\C\cap\R\PP^n\cap\{F_0'=0\}=\varnothing$.
\end{defn}
\begin{remarks}\em
(i) The condition $F_\C(\mathsf{H}^n_\infty(\C))\subset\mathsf{H}^m_\infty(\C)$ is equivalent to
$$
e:=\max\{\deg(f_1),\ldots,\deg(f_m)\}-\deg(f_0)>0.
$$

(ii) If the polynomials $f_0,f_1,\ldots,f_m$ are relatively prime, then
$$
Y_\C\cap\{F_0'=0\}=\{F_0'=0,F_1=0,\ldots,F_m=0\}.
$$

(ii) Quasi-polynomial maps include polynomial maps, but also more general regular maps. If $e>0$ and $\{F_0'=0\}\cap\R\PP^n=\varnothing$, then $f$ is a quasi polynomial map. This occurs for instance if $e>0$ and $f_0:=b_0^2+(b_1^2{\tt x}} \newcommand{\y}{{\tt y}_1^{2k}+\cdots+b_n^2{\tt x}} \newcommand{\y}{{\tt y}_n^{2k})^\ell$ where $k,\ell\geq0$ and $b_i\in\R\setminus\{0\}$.

(iv) If $n=1$, the condition $e:=\max\{\deg(f_1),\ldots,\deg(f_m)\}-\deg(f_0)>0$ characterizes quasi-polynomial maps.
\end{remarks}

\begin{thm}\label{property}
Let $S\subset\R^m$ be a semialgebraic set that is the image of a quasi-polynomial map $f:\R^n\to\R^m$. Then $S_{\infty}$ is connected.
\end{thm}
\begin{proof}[Proof of Theorem \em\ref{property} \em for the case $n=2$]
Write $f:=(\frac{f_1}{f_0},\cdots,\frac{f_m}{f_0}):\R^2\to\R^m$ where each $f_i\in\R[{\tt x}} \newcommand{\y}{{\tt y}_1,{\tt x}} \newcommand{\y}{{\tt y}_2]$ and $f_0$ has an empty zero set. Keep notations from \ref{s3a} and assume $\gcd(f_0,f_1,\ldots,f_m)=1$, so the set of points of indeterminacy of $F_{\mathbb K}} \newcommand{\N}{{\mathbb N}$ (where ${\mathbb K}} \newcommand{\N}{{\mathbb N}=\R$ or $\C$) is
$$
Y_{\mathbb K}} \newcommand{\N}{{\mathbb N}=\{x_0F_0'=0,F_1=0,\ldots,F_m=0\}\cap{\mathbb K}} \newcommand{\N}{{\mathbb N}\PP^2.
$$
By \ref{res} $Y_{\mathbb K}} \newcommand{\N}{{\mathbb N}$ is a finite subset of $\ell_\infty({\mathbb K}} \newcommand{\N}{{\mathbb N})\cup(\{F_0'=0\}\cap{\mathbb K}} \newcommand{\N}{{\mathbb N}\PP^2)$. As $f$ is quasi-polynomial, $Y_\R\subset\ell_\infty(\R)\setminus\{F_0'=0\}$. The proof is conducted in several steps:

\vspace{2mm}
\noindent{\bf Step 1}. \em Initial Preparation\em. Let $(Z_1,\pi_\C,\widehat{F}_\C)$ be an invariant resolution for $F_\C$. Keep notations from \ref{clue} and denote
\begin{itemize}
\item[$\bullet$] $X_1:=Z_1\cap\R\PP^k$, which is a non-singular real projective surface.
\item[$\bullet$] $\pi_\R:=\pi_\C|_{X_1}:X_1\to\R\PP^2$, which is the composition of a sequence of finitely many blow-ups and its restriction $\pi_\R|_{X_1\setminus\pi_\R^{-1}(Y_\R)}:X_1\setminus\pi_\R^{-1}(Y_\R)\to\R\PP^2\setminus Y_\R$ is a biregular isomorphism.
\item[$\bullet$] $\widehat{F}_\R:=\widehat{F}_\C|_{X_1}:X_1\to\R\PP^m$, which is a real regular map and satisfies
$$
\widehat{F}_\R|_{X_1\setminus\pi_\R^{-1}(Y_\R)}=F_\R\circ\pi_\R|_{X_1\setminus\pi_\R^{-1}(Y_\R)}.
$$
\end{itemize}
As $Y_\R\subset\ell_\infty(\R)$, it holds $\pi_\R(x)\in\R\PP^2\setminus\ell_\infty(\R)\equiv\R^2$ for each point $x\in X_1\setminus\pi_\R^{-1}(\ell_\infty(\R))$. Thus,
$$
\widehat{F}_\R(x)=F_\R\circ\pi_\R(x)=f(\pi_\R(x))\in\R^m\equiv\R\PP^m\setminus\mathsf{H}_\infty(\R),
$$
so $\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R))\subset\pi_\R^{-1}(\ell_\infty(\R))$.

\paragraph{}\label{comp}
The strict transform $\mathsf{K}_\infty$ under $\pi_\C$ of $\ell_\infty(\C)$ is an invariant non-singular rational curve and $\pi_\C|_{\mathsf{K}_\infty}:\mathsf{K}_\infty\to\ell_\infty(\C)\equiv\C\PP^1$ is an invariant biregular isomorphism. Consequently, $\mathsf{C}_\infty:=\mathsf{K}_\infty\cap\R\PP^k$ is a real non-singular rational curve and it is the strict transform under $\pi_\R$ of $\ell_\infty(\R)$.

\paragraph{}\label{E} Define $\mathsf{E}:=\{F_0'(z)=0\}\subset\C\PP^2$ and let $\widetilde{\mathsf{E}}$ be its strict transform under $\pi_\C$, which is an invariant projective curve. As $f_0$ has empty zero set and ${\tt x}} \newcommand{\y}{{\tt y}_0$ does not divide $F_0'$, the intersection $\mathsf{E}\cap\R\PP^2$ is a finite subset of $\ell_\infty(\R)$. In addition $\pi_\C^{-1}(\mathsf{E})=\widetilde{\mathsf{E}}\cup\bigcup_{y\in\mathsf{E}\cap Y_\C}\pi_\C^{-1}(y)$. We have:

\paragraph{}\label{E1} $\widetilde{\mathsf{E}}\cap\bigcup_{y\in Y_\R}\pi^{-1}_\C(y)=\varnothing$ and $\widetilde{\mathsf{E}}\cap\R\PP^k\subset\mathsf{C}_\infty$.

Let $y\in Y_\R$ and let us show $\widetilde{\mathsf{E}}\cap\pi_\C^{-1}(y)=\varnothing$. Otherwise, there exists $x\in\widetilde{\mathsf{E}}\cap\pi_\C^{-1}(y)$, so $y=\pi_\C(x)\in\pi_\C(\widetilde{\mathsf{E}})=\mathsf{E}$. Consequently, $y\in Y_\R\cap\mathsf{E}=Y_\R\cap\{F_0'=0\}=\varnothing$, which is a contradiction. As $\pi_\C(\widetilde{\mathsf{E}}\cap\R\PP^k)\subset\mathsf{E}\cap\R\PP^2\subset\ell_\infty(\R)$, we have
$$
\widetilde{\mathsf{E}}\cap\R\PP^k\subset\pi_{\R}^{-1}(\ell_\infty(\R))\setminus\bigcup_{y\in Y_\R}\pi_\R^{-1}(y)\subset\mathsf{C}_\infty.
$$

\paragraph{}\label{E2} As $\widehat{F}_\C|_{Z_1\setminus\pi_\C^{-1}(Y_\C)}=F_\C\circ\pi_\C|_{Z_1\setminus\pi_\C^{-1}(Y_\C)}$, it holds
$$
\widetilde{\mathsf{E}}\cup \mathsf{K}_\infty\subset \widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))\subset\widetilde{\mathsf{E}}\cup \mathsf{K}_\infty\cup\pi_\C^{-1}(Y_\C).
$$
Thus, $\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$ is an invariant projective curve whose irreducible components different from $\widetilde{\mathsf{E}}$ are by \ref{clue} either singletons or non-singular rational curves.

\paragraph{}\label{diagram} The following diagram summarizes the achieved situation:

\begin{center}
\begin{tikzpicture}
 \node (e) at (0,3.9) {$\widetilde{\mathsf{E}}\cup\mathsf{K}_\infty$};
 \node (c) at (0,3.4) {$\cap$};
 \node (f) at (-0.2,2.9) {$\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$};
 \node (c2) at (0,2.4) {$\cap$};
	 \node (z1) at (0,1.95) {$Z_1$};
	 \node (cp2) at (3,2) {$\C\PP^2$};
	 \node (c2) at (3,2.4) {$\cap$};
	 \node (hr) at (3,2.9) {$\ell_\infty(\C)\cup\mathsf{E}$};
	 \node (ss) at (4.125,2.92) {$\supset$};
	 \node (hr) at (4.6,2.9) {$Y_\C$};
	 \node (cpm) at (3,0) {$\C\PP^m$};
	 \node (cup) at (3,-0.45) {$\cup$};
	 \node (hr) at (3,-1) {$\mathsf{H}_\infty(\C)$};
\draw [->] (z1.east) to node [above] {$\pi_\C$} (cp2.west);
\draw [->] (z1.south east) to node [left] {$\widehat{F}_\C$} (cpm.north west);
\draw [->] (cp2.south) to node [right] {$F_\C$} (cpm.north);
\end{tikzpicture}
\begin{tikzpicture}
 \node (e) at (0,3.9) {$\widetilde{\mathsf{E}}\cap\R\PP^k$};
 \node (c) at (0,3.4) {$\cap$};
 \node (f) at (0,2.9) {$\mathsf{C}_\infty$};
 \node (c2) at (0,2.4) {$\cap$};
	 \node (z1) at (0,1.95) {$X_1$};
	 \node (cp2) at (3,2) {$\R\PP^2$};
	 \node (c2) at (3,2.4) {$\cap$};
	 \node (hr) at (3,2.9) {$\ell_\infty(\R)$};
	 \node (ss) at (3.75,2.92) {$\supset$};
	 \node (hr) at (4.25,2.9) {$Y_\R$};
	 \node (cpm) at (3,0) {$\R\PP^m$};
	 \node (cup) at (3,-0.45) {$\cup$};
	 \node (hr) at (3,-1) {$\mathsf{H}_\infty(\R)$};
\draw [->] (z1.east) to node [above] {$\pi_\R$} (cp2.west);
\draw [->] (z1.south east) to node [left] {$\widehat{F}_\R$} (cpm.north west);
\draw [->] (cp2.south) to node [right] {$F_\R$} (cpm.north);
\end{tikzpicture}
\end{center}

\noindent{\bf Step 2.} We prove next:

\paragraph{}\label{usrc}\em $\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$ is connected\em. Consequently, \em $\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$ does not contain isolated points and its irreducible components different from $\widetilde{\mathsf{E}}$ are non-singular rational curves\em.

Indeed, by Stein's factorization theorem \cite[III.11.5]{h2} applied to the projective morphism $\widehat{F}_\C:Z_1\to\C\PP^m$, there exist a projective variety $V$ and projective morphisms $G_1:Z_1\to V$ and $G_2:V\to\C\PP^m$ such that:
\begin{itemize}
\item[$\bullet$]$G_1$ is surjective and its fibers are connected,
\item[$\bullet$]$G_2$ is a finite morphism and
\item[$\bullet$]$\widehat{F}_\C=G_2\circ G_1$.
\end{itemize}
To prove \ref{usrc} it is enough to show that $\mathsf{H}:=G_2^{-1}(\mathsf{H}_\infty(\C))$ and $G_1^{-1}(\mathsf{H})=\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$ are connected.

\paragraph{} \em $\mathsf{H}$ is connected\em.

As $G_2$ is finite, it is by \cite[II.5.17]{h2} an affine morphism. Thus,
$$
G_2^{-1}(\C^m)=G_2^{-1}(\C\PP^m)\setminus G_2^{-1}(\mathsf{H}_\infty(\C))=V\setminus\mathsf{H}
$$
is an affine algebraic variety. As $V$ is a complete manifold, we deduce by \cite[6.2, p.79]{h} that $\mathsf{H}=V\setminus G_2^{-1}(\C^m)$ is connected because it is the complement in $V$ of an affine open subvariety.

\paragraph{} \em $G_1^{-1}(\mathsf{H})$ is connected\em.

Suppose that $G_1^{-1}(\mathsf{H})$ is the disjoint union of two closed subsets $A_1,A_2$. As $G_1$ is proper and surjective, $G_1(A_1),G_1(A_2)$ are closed subsets of $\mathsf{H}$ and $\mathsf{H}=G_1(A_1)\cup G_1(A_2)$. In case $G_1(A_i)\neq\varnothing$ for $i=1,2$, the intersection $G_1(A_1)\cap G_1(A_2)\neq\varnothing$ because $\mathsf{H}$ is connected. If $x\in G_1(A_1)\cap G_1(A_2)$, the fiber
$$
G_1^{-1}(\{x\})=(G_1^{-1}(\{x\})\cap A_1)\sqcup(G_1^{-1}(\{x\})\cap A_2)
$$
is the disjoint union of two non-empty closed sets, so $G_1^{-1}(\{x\})$ is disconnected, which is a contradiction because the fibers of $G_1$ are connected.

\vspace{2mm}
\noindent{\bf Step 3.} In the following we use Lemma \ref{crack} several times. To ease the procedure we point out the key facts. By \ref{curve2} $A:=\pi_\C^{-1}(\ell_\infty(\C)\cup Y_\C)$ is an algebraic curve whose irreducible components are non-singular rational curves. Observe that $U:=Z_1\setminus A=\pi_\C^{-1}(\C^2\setminus Y_\C)$ is a dense subset of $Z_1$ biregularly equivalent to $\C^2\setminus Y_\C$ (that is, the complement in $\C^2$ of a finite subset).

\paragraph{}\label{ucrack} \em Let $B_1,B_2\subset A$ be two connected compact algebraic curves without common irreducible components. Then the intersection $B_1\cap B_2$ is \em by Lemma \ref{crack}, Example \ref{bouquet} and Remark \ref{crackc} \em either empty or a singleton\em.

\vspace{2mm}
\noindent{\bf Step 4.} By Zariski's Main Theorem \cite[III.11.4]{h2} applied to the birational projective morphism $\pi_\C:Z_1\to\C\PP^2$, \em the fiber $\pi_\C^{-1}(y)$ is connected for each $y\in Y_\C$\em. We prove next:

\paragraph{}\label{connected2}\em If $y\in Y_\R$, the invariant projective variety $T_y:=\pi_\C^{-1}(y)\cap \widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$ is connected and $\mathsf{K}_\infty\cap T_y$ is a singleton\em. In addition, \em if $T_y$ has dimension $1$ and $\mathsf{K}_{1,y},\ldots,\mathsf{K}_{r_y,y}$ are the invariant irreducible components of $T_y$, the projective curve $\bigcup_{i=1}^{r_y}\mathsf{K}_{i,y}$ is connected and
$$
\mathsf{K}_\infty\cap T_y=\mathsf{K}_\infty\cap\bigcup_{i=1}^{r_y}\mathsf{K}_{i,y}
$$
is a singleton contained in $X_1$\em.

The clue to prove \ref{connected2} is the following:

\paragraph{}\label{connected0}\em Let $y\in Y_\R$ and $T$ be an invariant connected union of irreducible components of $\pi^{-1}_\C(y)$. Let $K_1,\ldots,K_r$ be the invariant irreducible components of $T$ and denote $C_i:=K_i\cap\R\PP^k$, which is \em by \ref{clue}(v) \em a real non-singular rational curve for $i=1,\ldots,r$. We have:
\begin{itemize}
\item[(i)] The projective curve $\bigcup_{i=1}^rK_i$ is connected.
\item[(ii)] Suppose moreover $\mathsf{K}_\infty\cap T\neq\varnothing$. Then the intersection $\mathsf{K}_\infty\cap T=\mathsf{K}_\infty\cap\bigcup_{i=1}^rK_i$ is a singleton contained in $X_1$. In particular, $\mathsf{K}_\infty\cup\bigcup_{i=1}^rK_i$ is connected.
\item[(iii)] We may order the indices $i=1,\ldots,r$ in such a way that $\mathsf{C}_\infty\cap C_1=\mathsf{K}_\infty\cap T$ is a singleton and $C_i\cap\bigcup_{j=1}^{i-1}C_j$ is a singleton for $i=2,\ldots,r$. In particular, the real projective curve $C:=\bigcup_{i=1}^rC_i$ is connected and $\mathsf{C}_\infty\cap C\neq\varnothing$.
\end{itemize}\em

We prove first \ref{connected0}(i). If $T=\bigcup_{i=1}^rK_i$, there is nothing to prove.

Otherwise, denote the non-invariant irreducible components of $T$ with $K_{r+1},\ldots,K_s$. Denote
$$
t:=\max\Big\{\#{\mathcal F}:\ {\mathcal F}\subset\{1,\ldots,r\},\ \bigcup_{i\in{\mathcal F}}K_i\quad\text{is connected}\Big\}
$$
and let us check $t=r$. Suppose by contradiction $t<r$. We may assume that $K:=\bigcup_{i=1}^tK_i$ is connected. As each $K_\ell$ is connected, each intersection $K\cap K_\ell=\varnothing$ for $t<\ell\leq r$. As $T$ is connected and invariant and $K\cap\bigcup_{i=t+1}^rK_i=\varnothing$, we may assume $K\cap K_{r+1}\neq\varnothing$ and $\sigma(K_{r+1})=K_{r+2}$. As $K$ is invariant, $K\cap K_{r+2}\neq\varnothing$, so $K\cup K_{r+1}\cup K_{r+2}$ is connected and invariant. Repeating the previous argument recursively, we find indices $r<r+2j_0\leq s$ and $t<\ell\leq r$ such that (after reordering the indices $i=r+1,\ldots,s$) $K_\ell\cap(K\cup\bigcup_{i=r+1}^{r+{2j_0}}K_j)\neq\varnothing$ and for each $1\leq j\leq j_0$ it holds
\begin{itemize}
\item[$\bullet$]$K\cup\bigcup_{i=r+1}^{r+{2j}}K_i$ is connected and invariant and
\item[$\bullet$]$\sigma(K_{r+2j-1})=K_{r+2j}$.
\end{itemize}
Such indices $r+2j_0,\ell$ exist because $T=\bigcup_{i=1}^sK_i$ is connected.

The invariant connected projective curves $K_\ell$ and $K\cup\bigcup_{i=r+1}^{r+{2j}}K_j$ are contained in $A:=\pi_\C^{-1}(\ell_\infty(\C)\cup Y_\C)$ because $y\in Y_\R$. The non-empty invariant intersection $K_\ell\cap(K\cup\bigcup_{i=r+1}^{r+{2j}}K_j)$ is by \ref{ucrack} a singleton $\{p_\ell\}\subset X_1=Z_1\cap\R\PP^k$. As $K_i\subset\C\PP^k\setminus\R\PP^k$ for $i=r+1,\ldots,s$ (see \ref{clue}(iv)), we have $p_\ell\in K_\ell\cap K\neq\varnothing$, which is a contradiction. Then $r=t$, so $\bigcup_{i=1}^rK_i$ is connected.

Next we prove \ref{connected0}(ii). Since $\mathsf{K}_\infty,T\subset A$ are invariant connected projective curves, the non-empty invariant intersection $\mathsf{K}_\infty\cap T$ is by \ref{ucrack} a singleton $\{p\}\subset X_1$. Thus, $p\in T\cap\R\PP^k\subset\bigcup_{i=1}^rK_i$ because the non-invariant irreducible components of $T$ are by \ref{clue}(v) contained in $\C\PP^k\setminus\R\PP^k$. Consequently, $\mathsf{K}_\infty\cap T=\mathsf{K}_\infty\cap\bigcup_{i=1}^rK_i=\{p\}$.

Finally, we show \ref{connected0}(iii). As $p\in X_1=Z_1\cap\R\PP^k$,
$$
\mathsf{C}_\infty\cap C=\mathsf{C}_\infty\cap\bigcup_{i=1}^rC_i=\Big(\mathsf{K}_\infty\cap\bigcup_{i=1}^rK_i\Big)\cap\R\PP^k=\{p\}\neq\varnothing.
$$
We may assume $\mathsf{C}_\infty\cap C_1\neq\varnothing$, that is, $\mathsf{C}_\infty\cap C_1=\{p\}=\mathsf{K}_\infty\cap T$. As $\bigcup_{i=1}^rK_r$ is connected, we claim: \em We may order the indices $i=2,\ldots,r$ in such a way that $K_i$ intersects the union $\bigcup_{j=1}^{i-1}K_j$ for $i=2,\ldots,r$\em.

Indeed, as $K=\bigcup_{i=1}^rK_i$ is connected, the intersection $K_1\cap\bigcup_{i=2}^rK_i\neq\varnothing$ and we assume $K_1\cap K_2\neq\varnothing$. As $K$ is connected, the intersection of $K_1\cup K_2$ and $\bigcup_{i=3}^rK_i\neq\varnothing$ is non-empty. We may assume that $K_3$ intersects $K_1\cup K_2$ and proceeding this way we prove the claim.

Next, since $K_i$ and $\bigcup_{j=1}^{i-1}K_j$ are invariant connected projective curves contained in $A$, the non-empty invariant intersection $K_i\cap\bigcup_{j=1}^{i-1}K_j$ is by \ref{ucrack} a singleton $\{q_i\}\subset X_1$. Thus,
$$
q_i\in\R\PP^k\cap K_i\cap\bigcup_{j=1}^{i-1}K_j=C_i\cap\bigcup_{j=1}^{i-1}C_j,
$$
so $C_i\cap\bigcup_{j=1}^{i-1}C_j\neq\varnothing$ for $i=2,\ldots,r$. As each $C_i$ is a non-singular curve biregularly equivalent to $\R\PP^1$, we deduce that the projective curve $C:=\bigcup_{i=1}^rC_i$ is connected.

\paragraph{}\label{connected2p} Now we are ready to prove \ref{connected2}. As $y\in Y_\R\subset\ell_{\infty}(\R)\subset\ell_{\infty}(\C)$ and $\pi_\C(\mathsf{K}_\infty)=\ell_{\infty}(\C)$, we deduce $\pi_\C^{-1}(y)\cap\mathsf{K}_\infty\neq\varnothing$. If $y_1\neq y_2$, the intersection $\pi_\C^{-1}(y_1)\cap\pi_\C^{-1}(y_2)=\varnothing$ and by \ref{E1} $\widetilde{\mathsf{E}}\cap\pi_\C^{-1}(y)=\varnothing$. Consequently, by \ref{E2}
\begin{multline*}
T_y\cap\mathsf{K}_\infty=\pi_\C^{-1}(y)\cap \widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))\cap\mathsf{K}_\infty\\
=((\pi^{-1}_\C(y)\cap\mathsf{K}_\infty)\cup\pi^{-1}_\C(y))\cap\mathsf{K}_\infty=\pi^{-1}_\C(y)\cap\mathsf{K}_\infty\neq\varnothing,
\end{multline*}
so by \ref{connected0}(ii) $T_y\cap\mathsf{K}_\infty$ is a singleton. Denote
$$
R_y:=\widetilde{\mathsf{E}}\cup\mathsf{K}_\infty\cup\bigsqcup_{z\in Y_\C,z\neq y}(\pi^{-1}_\C(z)\cap\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C)))
$$
and observe $\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))=T_y\cup R_y$ by \ref{E2}. Using again $\widetilde{\mathsf{E}}\cap\pi_\C^{-1}(y)=\varnothing$ and $\pi_\C^{-1}(y_1)\cap\pi_\C^{-1}(y_2)=\varnothing$ if $y_1\neq y_2$, we deduce $T_y\cap R_y=T_y\cap\mathsf{K}_\infty$, which is a singleton. Consequently, if $T_y$ was disconnected, then $\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$ would have been disconnected, which contradicts \ref{usrc}. Thus, the first part of claim \ref{connected2} holds. The second part follows readily from \ref{connected0}.

\vspace{2mm}
\noindent{\bf Step 5.} Next we show:

\paragraph{}\label{connected}\em $\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R))$ is connected\em. Consequently, \em $\widehat{F}_\R(\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R)))$ is also connected\em.

As $\widehat{F}_\C$ is invariant, $\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R))=\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))\cap\R\PP^k$. By \ref{E2}
\begin{equation}\label{star}
\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))=\widetilde{\mathsf{E}}\cup\mathsf{K}_\infty\cup\bigcup_{y\in Y_\C}(\pi^{-1}_\C(y)\cap \widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))).
\end{equation}
\indent Fix $y\in Y_\R$ and consider $T_y:=\pi_\C^{-1}(y)\cap \widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))$. If $T_y$ is a singleton, we deduce by \ref{connected2} that $T_y\subset\mathsf{K}_\infty$. Otherwise we denote the invariant irreducible components of $T_y$ with $\mathsf{K}_{1,y},\ldots,\mathsf{K}_{r_y,y}$. As the non-invariant irreducible components of $\pi_\C^{-1}(y)$ do not intersect $\R\PP^k$ by \ref{clue}(v), we deduce $T_y\cap\R\PP^k=\bigcup_{i=1}^{r_y}\mathsf{K}_{i,y}\cap\R\PP^k$.

In addition by \ref{clue}(iv) $\pi_\C^{-1}(y)\cap\R\PP^k=\varnothing$ for all $y\in Y_\C\setminus Y_\R$. Denote the subset of points of $Y_\R$ such that $T_y$ is not a singleton with $Y_\R'$. If we intersect expression \eqref{star} with $\R\PP^k$, we deduce by \ref{E1}
$$
\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R))=\widehat{F}_\C^{-1}(\mathsf{H}_\infty(\C))\cap\R\PP^k=\mathsf{C}_\infty\cup\bigcup_{y\in Y_\R'}\bigcup_{i=1}^{r_y}\mathsf{K}_{i,y}\cap\R\PP^k.
$$
\indent By \ref{connected2} the projective curve $\bigcup_{i=1}^{r_y}\mathsf{K}_{i,y}$ is connected and the intersection $\mathsf{K}_\infty\cap\bigcup_{i=1}^{r_y}\mathsf{K}_{i,y}$ is a singleton $\{p_y\}$ for each $y\in Y_\R'$. By \ref{connected0}(iii) each projective curve $\mathsf{C}_y:=\bigcup_{i=1}^{r_y}\mathsf{C}_{i,y}$ is connected and each intersection $\mathsf{C}_\infty\cap\mathsf{C}_y\neq\varnothing$. As $\mathsf{C}_\infty=\mathsf{K}_\infty\cap\R\PP^k$ is by \ref{comp} connected, we conclude that
$$
\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R))=\mathsf{C}_\infty\cup\bigcup_{y\in Y_\R'}\bigcup_{i=1}^{r_y}\mathsf{C}_{i,y}
$$
is connected too.

\vspace{2mm}
\noindent{\bf Final Step.} {\em Conclusion.} As $\widehat{F}_\R(\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R)))$ is connected, to prove that $S_\infty$ is connected, too, it is enough to show that both sets are equal, that is,
$$
\widehat{F}_\R(\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R)))=S_\infty.
$$
As $X_1=Z_1\cap\R\PP^k$ is compact, $\widehat{F}_\R$ is proper, so the restriction
$$
\widehat{F}_\R|_{\widehat{F}_\R^{-1}(\R^m)}:\widehat{F}_\R^{-1}(\R^m)\to\R^m
$$
is also proper. As $\pi^{-1}_\R(\R^2)$ is dense in $X_1$ and $S=f(\R^2)=F_\R(\R^2)=\widehat{F}_\R(\pi_\R^{-1}(\R^2))$, we have
\begin{multline*}
\widehat{F}_\R(\widehat{F}_\R^{-1}(\R^m))=\widehat{F}_\R(\operatorname{Cl}_{\widehat{F}_\R^{-1}(\R^m)}(\pi_\R^{-1}(\R^2)))\\
=\operatorname{Cl}_{\R^m}(\widehat{F}_\R(\pi_\R^{-1}(\R^2)))=\operatorname{Cl}_{\R^m}(f(\R^2))=\operatorname{Cl}_{\R^m}(S),
\end{multline*}
so $\widehat{F}_\R(\widehat{F}_\R^{-1}(\R^m))=\operatorname{Cl}_{\R^m}(S)$. As $\widehat{F}_\R$ is proper,
\begin{multline*}
\operatorname{Cl}_{\R^m}(S)\sqcup S_\infty=\operatorname{Cl}_{\R\PP^m}(S)=\operatorname{Cl}_{\R\PP^m}(\widehat{F}_\R(\widehat{F}_\R^{-1}(\R^m)))\\
=\widehat{F}_\R(\operatorname{Cl}_{X_1}(\widehat{F}_\R^{-1}(\R^m)))=\widehat{F}_\R(X_1)=\widehat{F}_\R(\widehat{F}_\R^{-1}(\R^m))\sqcup \widehat{F}_\R(\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R))).
\end{multline*}
Consequenlty, $S_\infty=\widehat{F}_\R(\widehat{F}_\R^{-1}(\mathsf{H}_\infty(\R)))$, which is by \ref{connected} connected, as required.
\end{proof}

We are ready to prove Theorem \ref{main1}. We present an independent proof from the one of Theorem \ref{property}. This is enlightening for the proof of Theorem \ref{property} for $n\geq3$.

\begin{proof}[Proof of Theorem \em\ref{main1}]
For $n=1$ the result follows from \cite[1.1]{fe}. To prove that $S_{\infty}$ is connected if $n\geq2$, it is enough to show that for any given pair of points $p,q\in S_{\infty} $ there exists a connected subset of $S_\infty$ containing $p$ and $q$. By Lemma \ref{polcurve} there exist polynomials ${\tt p}_i,{\tt q}_i\in\R[\t]$ such that ${\tt p}_i(0),{\tt q}_i(0)\neq0$ and integers $k_i,\ell_i$ such that the rational paths $\alpha:=(\t^{k_1}{\tt p}_1,\ldots,\t^{k_n}{\tt p}_n)$ and $\beta:=(\t^{\ell_1}{\tt q}_1,\ldots,\t^{\ell_n}{\tt q}_n)$ satisfy
$$
\lim_{t\to0^+}(f\circ\alpha)(t)=p\quad\text{and}\quad\lim_{t\to0^+}(f\circ\beta)(t)=q.
$$
At least one couple $(k_i,\ell_j)$ is of negative integers. Consider the polynomials
$$
P_i({\tt x}} \newcommand{\y}{{\tt y},\y):=\left\{\begin{array}{cl}
\y^{|k_i|}{\tt p}_i({\tt x}} \newcommand{\y}{{\tt y})&\text{if $k_i<0$,}\\[4pt]
{\tt x}} \newcommand{\y}{{\tt y}^{k_i}{\tt p}_i({\tt x}} \newcommand{\y}{{\tt y})&\text{if $k_i\geq0$}
\end{array}\right.
\quad\text{and}\quad
Q_i({\tt x}} \newcommand{\y}{{\tt y},\y):=\left\{\begin{array}{cl}
(-\y)^{|\ell_i|}{\tt q}_i({\tt x}} \newcommand{\y}{{\tt y})&\text{if $\ell_i<0$,}\\[4pt]
{\tt x}} \newcommand{\y}{{\tt y}^{\ell_i}{\tt q}_i({\tt x}} \newcommand{\y}{{\tt y})&\text{if $\ell_i\geq0$}
\end{array}\right.
$$
and let
$$
h:=\frac{{\tt x}} \newcommand{\y}{{\tt y}\y+1}{2}(P_1,\ldots,P_n)+\frac{1-{\tt x}} \newcommand{\y}{{\tt y}\y}{2}(Q_1,\ldots,Q_n).
$$
Consider the polynomial map $g:=f\circ h:\R^2\to\R^m$ and observe
\begin{itemize}
\item[$\bullet$]$T_0:=\operatorname{im}(g)\subset\operatorname{im}(f)=S$, so $T_{0,\infty}\subset S_{\infty}$.
\item[$\bullet$]$p=\lim_{t\to0^+}g(t,\frac{1}{t})\in T_{0,\infty}$ and $q=\lim_{t\to0^+}g(t,-\frac{1}{t})\in T_{0,\infty}$.
\end{itemize}
As $T_{0,\infty}$ is connected for $n=2$ by Theorem \ref{property}, we are done.
\end{proof}

\subsection{Proof of Theorem \ref{property}}\setcounter{paragraph}{0}
Now we prove Theorem \ref{property} for an arbitrary $n$.

\begin{proof}[Proof of Theorem \em \ref{property} \em for an arbitrary $n$.]
The case $n=1$ follows from \cite[1.4]{fe}. Assume $n\geq2$ and write $f:=(\frac{f_1}{f_0},\ldots,\frac{f_m}{f_0})$ where each $f_j\in\R[{\tt x}} \newcommand{\y}{{\tt y}]$, $\gcd(f_0,f_1,\ldots,f_m)=1$ and $f_0$ does not vanish on $\R^n$. The proof is conducted in several steps:

\paragraph{} \em Initial assumptions to simplify the proof\em.
Assume $\deg(f_1)\geq\deg(f_j)$ for $j=1,\ldots,m$. After a change of the type $(y_1,\ldots,y_m)\mapsto(y_1,y_2+b_2y_1,\ldots,y_m+b_my_1)$ where $b_j\in\R$ we can suppose
$$
\deg(f_1)=\cdots=\deg(f_m)=d>\deg(f_0)=d-e
$$
for some $e\geq1$. Denote
$$
F_j:={\tt x}} \newcommand{\y}{{\tt y}_0^df_j\Big(\frac{{\tt x}} \newcommand{\y}{{\tt y}_1}{{\tt x}} \newcommand{\y}{{\tt y}_0},\ldots,\frac{{\tt x}} \newcommand{\y}{{\tt y}_n}{{\tt x}} \newcommand{\y}{{\tt y}_0}\Big)
$$
and $F_0={\tt x}} \newcommand{\y}{{\tt y}_0^eF_0'$ where $e\geq1$ and $F_0'\in\R[{\tt x}} \newcommand{\y}{{\tt y}_0,{\tt x}} \newcommand{\y}{{\tt y}_1,\ldots,{\tt x}} \newcommand{\y}{{\tt y}_n]$ is not divisible by ${\tt x}} \newcommand{\y}{{\tt y}_0$. Notice that ${\tt x}} \newcommand{\y}{{\tt y}_0$ does not divide $F_j$ for $j=1,\ldots,m$ because $\deg(f_j)=\deg(F_j)=d$. After a change of the type $(x_1,\ldots,x_m)\mapsto(x_1,x_2+a_2x_1,\ldots,x_n+a_nx_1)$ where $a_i\in\R$ we can suppose $\deg(f_j)=\deg_{{\tt x}} \newcommand{\y}{{\tt y}_1}(f_j)$ for each $j$.

\paragraph{} \em Equivalent formulation for the statement\em.
To prove that $S_{\infty}$ is connected, it is equivalent to show: \em There exists a point $p_0\in S_{\infty}$ such that for any other point $q\in S_{\infty}$ there exists a connected subset of $S_\infty$ containing $p_0$ and $q$\em.

We will prove this last fact. Fix
$$
p_0:=\lim_{t\to0^+}(F_0:F_1:\cdots:F_m)\Big(1:\frac{1}{t}:0:\cdots:0\Big)\in\mathsf{H}_\infty(\R)
$$
and let $q\in S_{\infty} $. By Lemma \ref{polcurve} there exist a rational path $$
\alpha:=(\t^{k_1}{\tt p}_1,\ldots,\t^{k_n}{\tt p}_n)\in\R(\t)^n
$$
where
\begin{itemize}
\item[$\bullet$]$k_i\in{\mathbb Z}} \newcommand{\R}{{\mathbb R}$, $k_{i_0}:=\min\{k_1,\ldots,k_n\}<0$,
\item[$\bullet$]${\tt p}_i\in\R[\t]$ with ${\tt p}_i(0)\neq 0$ for $i=1,\ldots,n$ and ${\tt p}_{i_0}=\pm1$
\end{itemize}
and an integer $r\geq1$ such that $q=\lim_{t\to0^+}(f\circ(\alpha+\beta))(t)$ for each $\beta\in(\t)^r\R[\t]^n$. After the change $\t\to\t^6$ we may assume $k_{i_0}\leq-6$ and even. We keep all initial notations.

\paragraph{} \em Construction of an auxiliary regular map\em.
Write ${\tt p}_i:=\sum_{j=0}^{d_i}a_{ij}\t^j$ where $d_i:=\deg({\tt p}_i)$ and consider the formula
$$
P_i({\tt x}} \newcommand{\y}{{\tt y},\y):=\sum_{j+k_i<0}a_{ij}\y^{|k_i+j|}+\sum_{j+k_i\geq0}a_{ij}\frac{{\tt x}} \newcommand{\y}{{\tt y}^{e_j}\y}{(({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^{q_j}}
$$
where $j+k_i+1=4q_j+e_j$, $q_j\geq0$ and $0\leq e_j\leq3$ for each $0\leq j\leq d_i$ such that $j+k_i\geq0$. Observe $P_i(\t,\frac{1}{\t})=\t^{k_i}{\tt p}_i(\t)$ for $i=1,\ldots,n$ and $P_{i_0}={\rm p}_{i_0}\y^{|k_{i_0}|}=\pm\y^{|k_{i_0}|}$.

Denote $\ell_0:=\max_i\{q_{d_i}:\ d_i+k_i\geq0\}$ and notice $(({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell P_i\in\R[{\tt x}} \newcommand{\y}{{\tt y},\y]$ for each $\ell\geq\ell_0$. In addition
\begin{multline*}
\deg((({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell P_i)\leq\max\{-k_i+4\ell,4\ell-4q_j+e_j+1:\ k_i+j\geq0\}\\
\leq 4\ell+\max\{-k_i,4\}\leq4\ell+|k_{i_0}|
\end{multline*}
and the equality $\deg((({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell P_i)=4\ell+|k_{i_0}|$ holds if and only if $k_i=k_{i_0}$. As $|k_{i_0}|\geq6$,
$$
\deg_{\tt x}} \newcommand{\y}{{\tt y}((({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell P_i)\leq
\left\{\!\!\begin{array}{ll}
2(\ell-q_{d_i})+3&\text{if $k_i+d_i\geq0$}\\[4pt]
2\ell&\text{if $k_i+d_i<0$}
\end{array}\!\!\right\}
< 2\ell+4<4\ell+|k_{i_0}|
$$
for all $i=1,\ldots,n$. Let $\ell:=\max\{\ell_0,r\}$ and define
$$
h_i({\tt x}} \newcommand{\y}{{\tt y},\y):=\left\{\begin{array}{ll}
(({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell&\text{if $i=0$,}\\[4pt]
(({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell P_1({\tt x}} \newcommand{\y}{{\tt y},\y)+{\rm p}_1(0){\tt x}} \newcommand{\y}{{\tt y}^{4\ell+|k_{i_0}|}&\text{if $i=1$,}\\[4pt]
(({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell P_i({\tt x}} \newcommand{\y}{{\tt y},\y)&\text{if $i=2,\ldots,n$.}
\end{array}\right.
$$
Consider the regular map $h:=(\frac{h_1}{h_0},\ldots,\frac{h_n}{h_0}):\R^2\to\R^n$ and observe
\begin{multline*}
h(\t,0)=({\rm p}_1(0)\t^{4\ell+|k_{i_0}|},0,\ldots,0)\quad
\text{and}\quad h\Big(\t,\frac{1}{\t}\Big)=\alpha(\t)+({\rm p}_1(0)\t^{8\ell+|k_{i_0}|},0,\ldots,0),
\end{multline*}
so $\lim_{t\to0^+}(f\circ h)(\frac{1}{t},0)=p_0$ and $\lim_{t\to0^+}(f\circ h)(t,\frac{1}{t})=q$.

\paragraph{} \em Construction of an auxiliary quasi-polynomial map such that $p_0,q$ belong to the set of points at infinity of its image\em.
Let $g:=f\circ h:\R^2\to\R^m$ and denote $g_i:=F_i(h_0,h_1,\ldots,h_n)\in\R[\u_1,\u_2]$ for $i=0,1,\ldots,m$. Observe $g=(\frac{g_1}{g_0},\ldots,\frac{g_m}{g_0})$ and $g_0$ does not vanish on $\R^2$. We claim: \em $g$ is a quasi-polynomial map and $p_0,q\in g(\R^2)_{\infty}$\em.

First we have
\begin{itemize}
\item[$\bullet$] $T_0:=\operatorname{im}(g)\subset\operatorname{im}(f)=S$, so $T_{0,\infty}\subset S_{\infty}$.
\item[$\bullet$] $p_0=\lim_{t\to0^+}g(\frac{1}{t},0)\in T_{0,\infty}$ and $q=\lim_{t\to0^+}g(t,\frac{1}{t})\in T_{0,\infty}$.
\item[$\bullet$] $g_0=((\u_1\u_2-1)^2+\u_2^4)^{e\ell}F_0'(h_0,h_1,\ldots,h_n)$.
\end{itemize}
\noindent In addition,
$$
\deg(g_0)=d(4\ell+|k_{i_0}|)-e|k_{i_0}|<d(4\ell+|k_{i_0}|)=\max\{\deg(g_1),\ldots,\deg(g_m)\}.
$$

Indeed, since
\begin{itemize}
\item[$\bullet$] $\deg(F_0')=\deg(f_0)=\deg_{{\tt x}} \newcommand{\y}{{\tt y}_1}(f_0)=d-e$,
\item[$\bullet$] $\deg(F_j)=\deg(f_j)=\deg_{{\tt x}} \newcommand{\y}{{\tt y}_1}(f_j)=d$,
\item[$\bullet$] $\deg(h_i)\leq 4\ell+|k_{i_0}|,\ \deg(h_1)=4\ell+|k_{i_0}|$ and
\item[$\bullet$] $(h_0,h_1,\ldots,h_n)(\t,0)=(1,{\rm p}_1(0)\t^{4\ell+|k_{i_0}|},0,\ldots,0)$,
\end{itemize}
we have
$$
\deg(g_0)=4e\ell+(d-e)(4\ell+|k_{i_0}|)=d(4\ell+|k_{i_0}|)-e|k_{i_0}|<d(4\ell+|k_{i_0}|)=\deg(g_j)
$$
for each $j=1,\ldots,m$.

Let $\mu:=4\ell+|k_{i_0}|$ and write $H_i:=\u_0^\mu h_i(\frac{\u_1}{\u_0},\frac{\u_2}{\u_0})$ and $G_j:=\u_0^{d\mu}g_j(\frac{\u_1}{\u_0},\frac{\u_2}{\u_0})$, which are homogeneous polynomials. Notice that
\begin{multline*}
G_j=\u_0^{d\mu}F_j\Big(h_0\Big(\frac{\u_1}{\u_0},\frac{\u_2}{\u_0}\Big),h_1\Big(\frac{\u_1}{\u_0},\frac{\u_2}{\u_0}\Big),\ldots,h_n\Big(\frac{\u_1}{\u_0},\frac{\u_2}{\u_0}\Big)\Big)=F_j(H_0,H_1,\ldots,H_n)\\[-15pt]
\end{multline*}
and $G_0=\u_0^{e|k_{i_0}|}G_0'$ where $G_0':=((\u_1\u_2-\u_0^2)^2+\u_2^4)^{e\ell}F_0'(H_0,H_1,\ldots,H_n)$. Counting degrees one realizes: \em $\u_0$ does not divide $G_0'$\em.

In the following all zero sets are considered in $\R\PP^n$. To prove that $g$ is quasi-polynomial it only remains to check $\{G_0'=0,G_1=0,\ldots,G_m=0\}=\varnothing$.

Indeed, as $\{F_0'=0,F_1=0,\ldots,F_n=0\}=\varnothing$, the following equality holds
\begin{multline*}
\{G_0'=0,G_1=0,\ldots,G_m=0\}=\{H_0=0,\ldots,H_n=0\}\\
\cup\{((\u_1\u_2-\u_0^2)^2+\u_2^4)^{e\ell}=0,G_1=0,\ldots,G_m=0\}.
\end{multline*}
Notice
\begin{align*}
H_0(\u_0,\u_1,\u_2)&=u_0^{|k_{i_0}|}((\u_1\u_2-\u_0^2)^2+\u_2^4)^\ell,\\
H_1(0,\u_1,\u_2)&=\begin{cases}
{\rm p_1}(0)\u_1^{4\ell+|k_{i_0}|}&\text{if $k_{i0}<k_1$},\\[4pt]
{\rm p_1}(0)\left((\u_1^2\u_2^2+\u_2^4)^\ell\u_2^{|k_{i_0}|}+\u_1^{4\ell+|k_{i_0}|}\right)&\text{if $k_{i0}=k_1$},
\end{cases}\\
H_i(0,0,\u_2)&=\begin{cases}
0&\text{if $i\geq2$ and $k_{i0}<k_i$},\\[4pt]
{\rm p_i}(0)\u_2^{4\ell+|k_{i_0}|}&\text{if $i\geq 2$ and $k_{i0}=k_i$}.
\end{cases}
\end{align*}
Consequently, $\{H_0=0,\ldots,H_n=0\}=\varnothing$ because
\begin{itemize}
\item[$\bullet$] $H_0=0$ provides $\u_0=0$,
\item[$\bullet$] $H_1(0,\u_1,\u_2)=0$ provides $\u_1=0$ (we have used here that $k_{i_0}$ is even) and
\item[$\bullet$] $H_{i_0}(0,0,\u_2)=0$ provides $\u_2=0$.
\end{itemize}
On the other hand $((\u_1\u_2-\u_0^2)^2+\u_2^4)^{e\ell}=0$ provides $\u_0=0,\u_2=0$. As $G_1=0$, we have
\begin{multline*}
0=G_1(0,\u_1,0)\\
=F_1(H_0(0,\u_1,0),H_1(0,\u_1,0),\ldots,H_n(0,\u_1,0))=F_1(0,\u_1^\mu,0,\ldots,0).
\end{multline*}
For the last equality use that $\deg_{\tt x}} \newcommand{\y}{{\tt y}((({\tt x}} \newcommand{\y}{{\tt y}\y-1)^2+\y^4)^\ell P_i)<4\ell+|k_{i_0}|$ for all $i=1,\ldots,n$. As $\deg(F_1)=\deg(f_1)=\deg_{{\tt x}} \newcommand{\y}{{\tt y}_1}(f_1)=\deg_{{\tt x}} \newcommand{\y}{{\tt y}_1}(F_1)$, we obtain $a:=F_1(0,1,0,\ldots,0)\neq0$. As
$$
0=F_1(0,\u_1^\mu,0,\ldots,0)=a\u_1^{d\mu},
$$
we get $\u_1=0$, so $\{((\u_1\u_2-\u_0^2)^2+\u_2^4)^{e\ell}=0,G_1=0,\ldots,G_m=0\}=\varnothing$. Therefore
$$
\{G_0'=0,G_1=0,\ldots,G_m=0\}=\varnothing,
$$
so $g:\R^2\to\R^m$ is a quasi-polynomial map.

\paragraph{}\em Conclusion\em.
By Theorem \ref{property} for the case $n=2$ (already proved in \ref{s3a}) applied to the quasi-polynomial map $g$, we deduce that $T_{0,\infty}=(g(\R^2))_\infty\subset S_\infty$ is connected and since $p_0,q\in T_{0,\infty}$, we are done.
\end{proof}

The set of points at infinity of a semialgebraic set $S\subset\R^m$ is a semialgebraic subset of the hyperplane of infinity $\mathsf{H}_\infty(\R)$ of $\R\PP^m$. It seems reasonable to ask the following.

\begin{squestion}\em
Let $\mathsf{S}_0$ be a connected closed semialgebraic subset of $\mathsf{H}_\infty(\R)$. \em Is there a polynomial (or a quasi-polynomial) map $f:\R^n\to\R^m$ such that $f(\R^n)_\infty=\mathsf{S}_0$?\em
\end{squestion}

For $m=2$ the answer is positive but for higher dimension we have no further information.
\begin{examples}\label{exxe}\em
For each connected closed semialgebraic subset $\mathsf{S}_0\subset\ell_\infty(\R)$ there exists a polynomial map $f:\R^2\to\R^2$ such that $\dim_\R(f(\R^2))=2$ and $(f(\R^2))_\infty=\mathsf{S}_0$.

If $\mathsf{S}_0$ is not a singleton, the assertion follows from \cite[1.2]{u2}. On the other hand, the polynomial map $f:\R^2\to\R^2,\ (x,y)\mapsto(x,y^2+x^2)$ satisfies $f(\R^2))_\infty=\{(0:1:0)\}$, which is a singleton.
\end{examples}

\begin{sremark}\em
We have introduced quasi-polynomial maps to understand the limit of the image of a regular map to have a connected set of points at infinity. The following examples show that they do not enjoy the desired behavior.

(i) The composition of the quasi-polynomial maps
\begin{multline*}
g:\R^2\to\R^2,\ (x,y)\mapsto(x,y^2)\\
\text{and}\quad f:\R^2\to\R^2,\ (x,y)\mapsto\Big(\frac{x^3}{1+x^2+y^2},\frac{y^3}{1+x^2+y^2}\Big)
\end{multline*}
is not a quasi-polynomial map.

(ii) The image of the quasi-polynomial map $g:\R\to\R^2,\ t\mapsto(\frac{1}{1+t^2},1+t^2)$ is the semialgebraic set $S=\{xy=1, y\geq1\}$, which is not a polynomial image of $\R^n$.
\end{sremark}

\section{Set of points at infinity of a regular image of $\R^n$}\label{s4}

We have proved in Section \ref{s3} that the set of points at infinity of the image of a quasi-polynomial map $f:\R^n\to\R^m$ is connected. This is no longer true in general for regular maps even if $n=1$.

\subsection{Preliminary examples}
We present some examples to illustrate the previous fact and to show that the conditions in the statement of Theorem \ref{property} are sharp.

\begin{examples}\label{ex1}\em
(i) The image of the regular map
$$
f:\R^2\to\R^2,\ (x,y)\mapsto\Big((xy-1)^2+x^2,\frac{1}{(xy-1)^2+x^2}\Big)
$$
is $S:=\{a>0,ab=1\}$, so $S_\infty=\{(0:1:0),(0:0:1)\}$ is disconnected.

(ii) The image of the regular map
$$
f:\R^2\to\R^2,\ (x,y)\mapsto\Big(\frac{x^2}{1+y^2},\frac{y^2}{1+x^2}\Big)
$$
is $S:=\{a\geq0,b\geq0,ab<1\}$, so $S_\infty=\{(0:1:0),(0:0:1)\}$ is disconnected. If we write $f:=(\frac{f_1}{f_0},\frac{f_2}{f_0})$ where each $f_i$ is a non-zero polynomial, then $\deg(f_0)\!=\max\{\deg(f_1),\deg(f_2)\}$.

(iii) The image of the regular map
$$
f:\R^2\to\R^2,\ (x,y)\mapsto\Big(\frac{(1+x^4)y^6}{(1+y^4)^2},\frac{(1+y^4)x^4}{(1+x^4)^3}\Big)
$$
is a semialgebraic set $S$ such that $S_\infty=\{(0:1:0),(0:0:1)\}$ is disconnected. If we write $f:=(\frac{f_1}{f_0},\frac{f_2}{f_0})$ where each $f_i$ is a non-zero polynomial, then $\deg(f_0)\!<\max\{\deg(f_1),\deg(f_2)\}$. The set $Y_\C=\{(0:1:0),(0:0:1)\}$ of indeterminacy of the rational map
$$
F_\C:=(F_0:F_1:F_2):\C\PP^2\dashrightarrow\C\PP^2
$$
is contained in $\{F_0'=0\}\cap\R\PP^2$ where $F_0'=({\tt x}} \newcommand{\y}{{\tt y}_0^4+{\tt x}} \newcommand{\y}{{\tt y}_1^4)^3({\tt x}} \newcommand{\y}{{\tt y}_0^4+{\tt x}} \newcommand{\y}{{\tt y}_2^4)^2$ and $F_0={\tt x}} \newcommand{\y}{{\tt y}_0^2F_0'$.
\begin{proof}
Observe that $f(\R^2)\subset\{a^2b\leq1,a\geq0,b\geq0\}$ because
$$
\Big(\frac{(1+x^4)y^6}{(1+y^4)^2}\Big)^2\Big(\frac{(1+y^4)x^4}{(1+x^4)^3}\Big)=\Big(\frac{y^4}{1+y^4}\Big)^3\Big(\frac{x^4}{1+x^4}\Big)\leq 1.
$$
Thus, $S_\infty\subset\{(0:1:0),(0:0:1)\}$. To prove the converse inclusion it is enough to pick two rational paths $\alpha_i:(0,1]\to\R^2$ such that $\lim_{t\to0^+}\|\alpha_i(t)\|^2=+\infty$ and
\begin{align*}
&\lim_{t\to0^+}\frac{g_1}{g_0}(\alpha_1(t))=+\infty,\ \lim_{t\to0^+}\frac{g_2}{g_0}(\alpha_1(t))=0,\\
&\lim_{t\to0^+}\frac{g_1}{g_0}(\alpha_2(t))=0,\ \lim_{t\to0^+}\frac{g_2}{g_0}(\alpha_2(t))=+\infty.
\end{align*}
For instance, $\alpha_1(\t):=(\frac{1}{\t},1)$ and $\alpha_2(t):=(1,\frac{1}{\t})$ do the job.
\end{proof}
\end{examples}

The following question arises naturally.

\begin{squestion}\em\label{q}
Given a closed semialgebraic subset $\mathsf{S}_0\subset\ell_\infty(\R)\subset\R\PP^2$: \em Is there a regular map $f:\R^2\to\R^2$ such that $(f(\R^2))_\infty=\mathsf{S}_0$?
\end{squestion}

In case $\mathsf{S}_0$ is either connected or a finite set, the answer is by Examples \ref{exxe} positive.

\subsection{More sophisticated examples}\setcounter{paragraph}{0}
If $\mathsf{S}_0$ is a finite set, we proceed as follows.

\begin{prop}
Let $H_i:=c_i{\tt x}} \newcommand{\y}{{\tt y}-d_i\y$ be linear equations such that $c_id_i\neq0$ and the lines $\ell_i:=\{H_i=0\}$ are pairwise different. Denote $p_i:=(0:d_i^2:c_i^2)$. Then the image of the regular map
$$
h:=(h_1,h_2):\R^2\to\R^2,\ (x,y)\mapsto\Big(\frac{x^2}{1+\prod_{i=1}^rH_i(x,y)^2},\frac{y^2}{1+\prod_{i=1}^rH_i(x,y)^2}\Big)
$$
is a semialgebraic set $S$ such that $S_\infty=\{p_1,\ldots,p_r\}$.
\end{prop}
\begin{proof}
Observe first
$$
h\Big(\frac{d_i}{t},\frac{c_i}{t}\Big)=\Big(\frac{d_i^2}{t^2},\frac{c_i^2}{t^2}\Big)\equiv(t^2:d_i^2:c_i^2)\stackrel{t\to 0^+}{\longrightarrow}(0:d_i^2:c_i^2)=p_i,
$$
so $\{p_1,\ldots,p_r\}\subset S_\infty$. Next, we prove the converse inclusion $S_\infty\subset\{p_1,\ldots,p_r\}$.

The Jacobian of $h$ is not identically zero, so $S:=h(\R^2)$ has dimension $2$. In addition, $S$ is pure dimensional because it is a regular image of $\R^2$. Thus,
$$
S':=S\setminus(h(\{x=0\})\cup h(\{y=0\})\cup h(\{\operatorname{Jac}(h)=0\}))
$$
is dense in $S$. Fix a point $p\in S_\infty$. After interchanging the variables and changing ${\tt x}} \newcommand{\y}{{\tt y}$ with $-{\tt x}} \newcommand{\y}{{\tt y}$ if necessary, there exists by Lemma \ref{polcurve} a rational path $\alpha:=(\t^{-k},\t^\ell{\rm p})$ where $k,\ell\in{\mathbb Z}} \newcommand{\R}{{\mathbb R}$, $-k=\min\{-k,\ell\}<0$, ${\rm p}\in\R[\t]$ and ${\rm p}(0)\neq0$ such that $\lim_{t\to0^+}(h\circ\alpha)=p$ and $(h\circ\alpha)((0,\varepsilon))\subset S'$ for $\varepsilon>0$ small enough. Our change of variables does not modify the structure of $h$, so we keep the same notations. As $p\in\ell_\infty(\R)$, one of the following limits is infinity:
\begin{align*}
\lim_{t\to0^{+}}(h_1\circ\alpha)(t)&=\lim_{t\to0^{+}}\frac{t^{2k(r-1)}}{t^{2kr}+\prod_{i=1}^rH_i(1, t^{k+\ell}{\rm p}(t))^2},\\[6pt]
\lim_{t\to0^{+}}(h_2\circ\alpha)(t)&=\lim_{t\to0^{+}}\frac{t^{2kr+2\ell}{\rm p}(t)^2}{t^{2kr}+\prod_{i=1}^rH_i(1,t^{k+\ell}{\rm p}(t))^2}.
\end{align*}
Notice the following:
\begin{itemize}
\item[(1)] The first limit is infinity if and only if
$$
t^{2kr}+\prod_{i=1}^rH_i(1,\t^{k+\ell}{\rm p}(\t))^2=\t^{\nu_1}{\rm q}_1(\t)
$$
for some ${\rm q}_1\in\R[\t]$ and an integer $\nu_1\geq 2k(r-1)+1$.
\item[(2)] The second limit is infinity if and only if $\ell<0$ and
$$
t^{2kr}+\prod_{i=1}^rH_i(1,\t^{k+\ell}{\rm p}(\t))^2=\t^{\nu_2}{\rm q}_2(\t)
$$
for some ${\rm q}_2\in\R[\t]$ and an integer $\nu_2\geq 2kr+2\ell+1$.
\end{itemize}
As $k+\ell\geq0$, we deduce in both cases that there exists an index $i=1,\ldots,r$ such that $\lim_{t\to0^+}H_i(1,t^{k+\ell}{\rm p}(t))=0$. Since by hypothesis $c_i,d_i\neq0$, we conclude $\ell=-k<0$ and ${\rm p}(0)=\frac{c_i}{d_i}$. Denote $\mu:=\nu_j-2k(r-1)-1\geq0$ and ${\rm q}:={\rm q}_j$ in both cases, so
$$
\t^{2kr}+\prod_{i=1}^rH_i(1,\t^{k+\ell}{\rm p}(\t))^2=\t^{2k(r-1)+1}\t^\mu{\rm q}.
$$
We deduce
\begin{equation*}
\begin{split}
p&=\lim_{t\to0^+}(t^{2kr}+\prod_{i=1}^rH_i(1,{\rm p}(t))^2:t^{2k(r-1)}:t^{2kr+2\ell}{\rm p}(t)^2)\\
&=\lim_{t\to0^+}(t^{2k(r-1)+1}t^\mu{\rm q}(t):t^{2k(r-1)}:t^{2k(r-1)}{\rm p}(t)^2)\\
&=\lim_{t\to0^+}(t^{\mu+1}{\rm q}(t):1:{\rm p}(t)^2)=\Big(0:1:\frac{c_i^2}{d_i^2}\Big),
\end{split}
\end{equation*}
so $S_\infty\subset\{p_1,\ldots,p_r\}$.
\end{proof}

We present next an example of a regular image $S$ such that $S_\infty$ has exactly two $1$-dimensional connected components.

\begin{lem}\label{2int}
There exists a regular map $f:\R^2\to\R^2$ whose image $S$ satisfies
$$
S_\infty=\Big\{(0:u:1):\ 0\leq u\leq \frac{1}{2}\Big\}\cup\Big\{(0:1:v):\ 0\leq v\leq \frac{1}{2}\Big\}.
$$
\end{lem}
\begin{proof}
We build $f$ as the composition of two regular maps that we construct next:

\paragraph{}
\em Let $T:=\{0<a\leq1,b>0\}\cup\{0<b\leq1,a>0\}$. The image of the regular map
$$
g:\R^2\to\R^2,\ (x,y)\mapsto\Big(\frac{x^2+1}{1+x^2y^2},\frac{y^2+1}{1+x^2y^2}\Big)
$$
is a semialgebraic set $S_1$ such that $T\subset S_1\subset T\cup[0,2]^2\subset\{a>0,b>0\}$\em. In particular, $S_{1,\infty}=\{(0:1:0),(0:0:1)\}$.

We check first $S_1\subset T\cup[0,2]^2$. Let $(x,y)\in \R^2$ and write $f_3(x,y)=:(a,b)$. We claim: \em If $b>2$, then $0<a\leq 1$\em. If $a>2$, we will have by symmetry $0<b\leq 1$, so $S_1\subset T\cup[0,2]^2$.

It is clear that $a>0$. Suppose by contradiction $a>1$. Then $x^2>x^2y^2$ and $y^2>1+2x^2y^2$, so
$$
x^2<x^2+2x^4y^2<x^2y^2<x^2,
$$
which is a contradiction.

Next, we check $T\subset S_1$. It is enough to prove by symmetry that $T_1:=\{0<a\leq1,b>0\}\subset S_1$. Let $(a,b)\in T_1$ and consider the system of equations
$$
\left\{\begin{array}{l}
x^2+1=a(1+x^2y^2),\\[4pt]
y^2+1=b(1+x^2y^2).
\end{array}\right.\quad\leadsto\quad
\left\{\begin{array}{l}
b(x^2+1)-a(y^2+1)=0,\\[4pt]
ay^4+(a-1-b)y^2+b-1=0.
\end{array}\right.
$$
A simple discussion shows that both systems are equivalent. The discriminant $\Delta$ of the biquadratic equation $ay^4+(a-1-b)y^2+b-1=0$ is
$$
(a-1-b)^2-4a(b-1)=(b-3a+1)^2+8a(1-a),
$$
which is $\geq0$ because $0<a\leq1$. As $a-1-b<0$, the real number
$$
z_0:=\frac{b+(1-a)+\sqrt{(b-3a+1)^2+8a(1-a)}}{2a}
$$
is positive and has a square root $y_0$, which is a solution of the biquadratic equation
$$
ay^4+(a-1-b)y^2+b-1=0.
$$
The equation $b(x^2+1)-a(y_0^2+1)=0$ has a real solution $x_0$ if
$$
0<2(a(y_0^2+1)-b)=2az_0+2a-2b=-b+1+a+\sqrt{(b-3a+1)^2+8a(1-a)}
$$
or equivalently if
$$
0<(b-3a+1)^2+8a(1-a)-(b-1-a)^2=4b(1-a).
$$
As $b>0$ and $a<1$, it holds $4b(1-a)>0$, so we deduce $(a,b)\in S_1$, as required.

\paragraph{}
Let $B_1:=\{0<2a\leq b\}$ and $B_2:=\{0<2b\leq a\}$. Write also $A_1:=\{0<x\leq 1,\ 4\leq y\}$ and $A_2:=\{0<y\leq 1,\ 4\leq x\}$. \em Then the image of $A:=A_1\cup A_2$ under the regular map
\begin{equation*}
\begin{split}
h:\R^2&\to\R^2,\\
(x,y)&\mapsto\Big(\frac{x((y-1)^2y^2+2(x-1)^2x)}{1+x(x-1)^2y(y-1)^2},\frac{y((x-1)^2x^2+2(y-1)^2y)}{1+x(x-1)^2y(y-1)^2}\Big),
\end{split}
\end{equation*}
is a semialgebraic set $S_2$ contained in $B:=B_1\cup B_2$, which satisfies
$$
S_{2,\infty}=B_{1,\infty}\cup B_{2,\infty}=\Big\{(0:u:1):\ 0\leq u\leq \frac{1}{2}\Big\}\cup\Big\{(0:1:v):\ 0\leq v\leq \frac{1}{2}\Big\}.
$$\em
Write $h:=(\frac{h_1}{h_0},\frac{h_2}{h_0})$ where
\begin{align*}
&h_0({\tt x}} \newcommand{\y}{{\tt y},\y):=1+{\tt x}} \newcommand{\y}{{\tt y}({\tt x}} \newcommand{\y}{{\tt y}-1)^2\y(\y-1)^2,\\
&h_1({\tt x}} \newcommand{\y}{{\tt y},\y):={\tt x}} \newcommand{\y}{{\tt y}((\y-1)^2\y^2+2({\tt x}} \newcommand{\y}{{\tt y}-1)^2{\tt x}} \newcommand{\y}{{\tt y}),\\
&h_2({\tt x}} \newcommand{\y}{{\tt y},\y):=\y(({\tt x}} \newcommand{\y}{{\tt y}-1)^2{\tt x}} \newcommand{\y}{{\tt y}^2+2(\y-1)^2\y).
\end{align*}
As $h_1(\y,{\tt x}} \newcommand{\y}{{\tt y})=h_2({\tt x}} \newcommand{\y}{{\tt y},\y)$ and $h_0(\y,{\tt x}} \newcommand{\y}{{\tt y})=h_0({\tt x}} \newcommand{\y}{{\tt y},\y)$, we have $\frac{h_1(\y,{\tt x}} \newcommand{\y}{{\tt y})}{h_0(\y,{\tt x}} \newcommand{\y}{{\tt y})}=\frac{h_2({\tt x}} \newcommand{\y}{{\tt y},\y)}{h_0({\tt x}} \newcommand{\y}{{\tt y},\y)}$, so it is enough to prove: \em $h(A_1)\subset B_1$ and $(h(A_1))_\infty=B_{1,\infty}$\em.

Let $(x,y)\in A_1$. It holds $h_1(x,y)>0$ and
\begin{multline*}
h_2(x,y)-2h_1(x,y)=y((x-1)^2x^2+2(y-1)^2y)-2(x((y-1)^2y^2+2(x-1)^2x))\\
=2(1-x)(y-1)^2y^2+(y-4)(x-1)^2x^2\geq0
\end{multline*}
because $0<x\leq 1$ and $y\geq 4$, so $h(x,y)\in B_1$. Therefore, $(h(A_1))_\infty\subset B_{1,\infty}$ and it only remains to check $B_{1,\infty}\subset(h(A_1))_\infty$.

Indeed, for each $0<\lambda\leq 1$ consider the half-line $x=\lambda,y=t\geq4$ and the curve $C_\lambda\subset h(A_1)$ parametrized by
\begin{multline*}
\alpha_\lambda(t):=(\alpha_{\lambda1}(t),\alpha_{\lambda2}(t)):=g(\lambda,t)
=\Big(\frac{\lambda(t-1)^2t^2+2\lambda\mu_\lambda)}{1+\mu_\lambda t(t-1)^2},\frac{2(t-1)^2t^2+\lambda\mu_\lambda t}{1+\mu_\lambda t(t-1)^2}\Big)\\[-15pt]
\end{multline*}
where $\mu_\lambda:=\lambda(\lambda-1)^2$. As
$$
\lim_{t\to+\infty}\alpha_{\lambda1}(t)=+\infty,\qquad\lim_{t\to+\infty}\alpha_{\lambda2}(t)=+\infty\qquad\text{and}\qquad\lim_{t\to+\infty}\frac{\alpha_{\lambda1}(t)}{\alpha_{\lambda2}(t)}=\frac{\lambda}{2},
$$
we deduce $C_{\lambda,\infty}=\{(0:\frac{\lambda}{2}:1)\}$, so
$$
B_{1,\infty}=\bigcup_{0<\lambda\leq 1}C_{\lambda,\infty}\subset\Big(\bigcup_{0<\lambda\leq 1}C_\lambda\Big)_\infty=(h(A_1))_\infty,
$$
as required.

\paragraph{}
The image of the regular map $f:=h\circ g:\R^2\to\R^2$ is a semialgebraic set $S$ such that $S_\infty=\{(0:u:1):\ 0\leq u\leq \frac{1}{2}\}\cup\{(0:1:v):\ 0\leq v\leq \frac{1}{2}\}$, as required.
\end{proof}

\begin{squestion}\em
Let $\mathsf{S}_0$ be a closed semialgebraic subset of the hyperplane of infinity $\mathsf{H}_\infty(\R)$ of $\R\PP^m$. \em Is there a regular map $f:\R^n\to\R^m$ such that $(f(\R^n))_\infty=\mathsf{S}_0$?
\end{squestion}

redpajama_set_name: RedPajamaArXiv


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Melinda's list: Selmys' forty ways to love a lesbian (and hate Michael Voris)
By Tom O'Toole
Homosexuality refers to relations between men or between women who experience an exclusive or predominant sexual attraction toward persons of the same sex. It has taken a great variety of forms through the centuries and in different cultures. Its psychological genesis remains largely unexplained. Basing itself on Sacred Scripture, which presents homosexual acts as acts of grave depravity, tradition has always declared that "homosexual acts are intrinsically disordered." They are contrary to the natural law. They close the sexual act to the gift of life. They do not proceed from a genuine affective and sexual complementarity. Under no circumstances can they be approved. (CCC #2357)
After being questioned by many (including myself) for condemning Michael Voris' documentary, "FBI (Faith Based Investigation) into Homosexuality," as erroneous, but then not giving any evidence to support this claim, former queer (her term) turned married Catholic mother of six, Melinda Selmys, went back to the blogging board and "called Voris' bluff," coming up with forty so-called errors in Voris' film. Sadly, many of these claims stretch the limits of credulity (if not veracity) and, as in the case of both of Selmys' previous two works on the subject ("How to Speak About Homosexuality" and "What I Mean By Acceptance"), raised more questions than they answered.
Part scholarly research and part knee-jerk reaction, Selmys herself admitted that many of Voris' "errors" are not actually errors at all, but merely her personal objections to Michael's style. For example, Error #6, "The entire 'history' section suffers from massive Western chauvanism. The result is a misleading and lop-sided portrayal," appears to be merely a typical feminist beef without fact, and #12, "Paranoia abounds. This is strictly speaking not a factual error, but it is an extremely problematic form of thinking," is even more problematic in that it harkens back to the original problem; Selmys is merely calling Voris names without giving reasons. As early as the third grade, students are now taught to give evidence to support their answers, and all parties concerned would have probably been better served if Selmys had made her list shorter and left such name calling off of it.
On the other hand, Selmys does give her reasons for the criticism of the scripture Voris uses, but one would be hard-pressed to find an orthodox Catholic who finds them reasonable. Of course, editing, both for context and balance, is a valid topic for the reviewer, and while many of Selmys' many suggestions for inclusion in this video seem valid, hers would have been a Lord of the Rings (uncut version) epic rather than the 94-minute documentary that Michael did. Still, her criticism "Scriptural quotations are consistently clipped in order to make them as harsh and condemning as possible, sometimes in a way that does damage to the actual text and its context," especially when she then cites "Romans 1, in particular," is peculiar. For as anyone familiar with Romans 1 can tell you, the longer you read Paul's rant against the gay Romans, the more "harsh and condemning" it becomes. "Mutual degradation of their bodies" and "unnatural relations" becomes "every form of...evil, greed and malice" and later "senseless, heartless, ruthless..." to name a few. So if Voris could be accused of anything in this case, it would be mercy rather than overkill.
But if Selmys' critique of Voris' use of "Romans" is curious, her problem with his reference to Genesis 19 – at least to the orthodox Catholic – is just crazy. "No discussion of the obvious fact," claims Selmys, "that the sexuality portrayed in Sodom was ritualized gang-rape of supernatural beings (angels), not consensual sex between human male individuals." Since, as Catholic Answers states, "[t]hroughout history, Jewish and Christian scholars have recognized that one of the chief sins involved in God's destruction of Sodom was its people's homosexual behavior," perhaps this novel interpretation of the passage can be attributed to the six or so homosexuals on the strange New Evangelization staff that Selmys is a part of and that Voris, not surprisingly is not. Notwithstanding the fact that the Church fathers have consistently taught the (main) sin of Sodom was homosexuality and not rape, if this "gang-rape" teaching was current Catholic theology, would not the Catholic Catechism referenced Genesis 19 in support of section #2356, which has to do with rape, rather than section #2357, which deals with homosexuality? Which makes me wonder what Selmys' real problem is anyway...
For when she critiques Voris' film, "It is strongly implied that Ancient Greece is the only historical society in which homosexual relations were socially accepted in order to develop the inference that this is highly abnormal. Actually, they were widely accepted during certain historical periods in Japan, China, certain parts of India, the Arab world (there was a particularly fertile period of homoerotic Muslim poetry and spiritually [sic] during the late Middle Ages), pre-colonial America...etc. etc," does she really disagree with his point that throughout history, unnatural sexual relationships were in the minority? Or is she merely making this list to make a plug for "homoerotic Muslim poetry," and those friends of hers who enjoy it? And when she complains, "Voris conflates 'sexual orienatation' [sic] with 'chosen sexual behaviour' in order to try to deny any legitimate affinity between the gay rights movement and the civil rights movement. This is very poor argumentation: it's a classic example of redefining the terms in mid-argument so that they will be better suited to making your point," is she really upset about Voris' choice of words, or does part of her think that this "affinity" is legit?
Surely Selmys knows that gays in America have all the same "rights" as straights with one exception; and those who now use this gay rights vs. civil rights argument have their eyes on that one prize, namely gay "marriage." But as Selmys professes to follow CCC #2357 that calls homosexual acts "intrinsically disordered," a "grave depravity," could she be arguing for gay couples living together in a chaste relationship? Certainly, this seems highly unlikely, given the fact that when she broke off her lesbian relationship to become Catholic and her girlfriend suggested it, Selmys replied this was "impractical" and "wouldn't work." Or is she throwing this diabolical bone of gay marriage to her non-Catholic friends in some Machiavellian attempt to one day win them over?
Selmys claimed in her rebuttal to my first article on her that I was suggesting that she should not write based on the fact she wasn't perfect. I replied my criticism was instead that while she was advertising herself as a Catholic, she, by mixing Catholic teaching and secular logic, was really playing with fire. Melinda, you argue with Voris' explanation that "'One can conclude that it's their [gays'] behaviour that's causing the problem,' the problem being elevated rates of mental illness in homosexuals" (from Voris Error #4) is simplistic and inconclusive. But did not St. Paul, who said that unnatural relations developed into senselessness and heartlessness, argue the same thing?
Could it be that your former "confused...suicidal, self-harming," self was caused to some, if not a large degree, by your homosexual relations? It is wonderful that you are a Catholic wife and mother now, and as a Catholic husband and father I couldn't be more happy that God has given you these graces. But as your "conservative" Catholic friend who likes to write a bit himself, I pray that, for the sake of your many devoted readers, your past demons (and present anger at Voris) do not begin to creep into your present apostolate and poison it forever.
Related: Speaking about homosexuality? Melinda Selmys' 'queer' way to save gays
© Tom O'Toole
Thomas Augustine O'Toole was born in Chicago and grew up in a devout Catholic family with five brothers and two sisters. He was the sports editor of Notre Dame's Scholastic magazine, where his story "Reflections on the Game" won the award for Best Sports Feature for the Indiana Collegiate Press Association... (more)
Thomas Augustine O'Toole was born in Chicago and grew up in a devout Catholic family with five brothers and two sisters. He was the sports editor of Notre Dame's Scholastic magazine, where his story "Reflections on the Game" won the award for Best Sports Feature for the Indiana Collegiate Press Association.
Besides his work as a columnist for several suburban newspapers, Tom O'Toole is also a popular freelance writer for many Catholic magazines, including Columbia, The Catholic Faith, Queen of All Hearts, and Culture Wars.
His innovative writings on the connection between faith and sports ("Cathletics") has been praised by the Papal Nuncio, and his groundbreaking book Champions of Faith: Catholic Sports Heroes Tell Their Stories is one of the original volumes to be placed on the Vatican shelves where the Pontifical Council of Church and Sport meet. In 2006, Tom published a number of articles on the prestigious Catholic Exchange website, and started his blog Fighting Irish Thomas, which has been praised for its Catholic candor and humor.
Tom O'Toole lives in the Chicago area with his wife Jeanette. They have four children. He can be reached at tacotoole@
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redpajama_set_name: RedPajamaCommonCrawl


The man accused of hitting a toddler on a Delta Air Lines flight from Minneapolis to Atlanta no longer has a job.
Al Haase, president and CEO of AGC Aerospace and Defense, Composites Group, said Joe Rickey Hundley was not with the company as of Sunday, according to the Associated Press. The AP reported Haase called the alleged behavior of the former company executive during a personal trip "disturbing."
Hundley, 60, of Hayden, ID, was charged with simple assault of 19-month-old Jonah Bennett after the Feb. 8 flight. The todler's is black; his adopted parents are white. The boy's mother, 33-year-old Jessica Bennett, said Hundley smelled of alcohol and "told her to shut that (N-word) baby up," court records showed. She told KARE Hundley made another racial slur and slapped the todler, scratching him below his right eye. She credited other passengers with coming to her aid and giving them another seat on the plane.
Hundley's employer initially suspended him after hearing of the alleged incident. A company spokesman would not say whether he resigned or was fired, only that the allegations were contradictory to the company's values. Hundley's attorney said he plans to plead not guilty and asked the public not to rush to judgment, according to CNN.
1-Usually if the parent is present, then the parent is left to do the disciplining.
2-The disciplining is out of love and corrective, usually addressing the violation of a familial or social value.
3-The disciplining party is sober and acting on principle.
4-The disciplining party is viewed as a member of our family or tight-knit community and not a total stranger, especially if the discipline is corporal.
5-The punishment must suit the crime.
I understand times have changed, unfortunately the need for disciplining children remains a cultural necessity. It is part of the human becoming. But in the above story none of the aforementioned elements of disciplining loved ones is applicable.
1-A parent was present and handling the situation.
2-The disciplining party acted out of frustration—his peace was being disturbed.
3-The disciplining party was inebriated.
4-The disciplining party was a total stranger.
5-Slapping a toddler in the face for crying is inappropriate and unacceptable. Toddlers cry.
An additional things make this incident especially heinous: The repeated racial slurs.
Should Mr Hundley have been fired, I don't know? If there are legal grounds for it then he should have been. If he violated company policy, sure. In a litigious society like America, he might be able to beat the case. But he broke the law and should definitely see some jail time for assault, a hate-crime, and a few other charges. A year is not long enough.
And he should have got that can he deserved and I'm not talking about a beer. Apparently he already had some of those cans and they helped, along with his racism, to get him into this mess.
Obama said that these types of people cling to their alcohol and guns. He forgot to mention their racism. It's a good thing this time he only had his racism and alcohol. Who knows what he might have done if he had his gun too.
I so agree with you Seba!! Can you imagine if this drunken racist had a gun? That baby would be dead!!! JUST LIKE THAT!!
This is totally unacceptable. I cannot believe that he had to audacity to slap a toddler. I try to imagine if the roles were reversed and a black man slapped a white toddler. I know all hell would break loose. It is only fair to get equal treatment and try as much as possible to stop racism. I really wonder when we will sober up and live as one nation. This is way too much and totally off limits.


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