text:
\section{Introduction}\label{intro}
\vskip 5pt

\subsection{Topological Posets}\label{toppo}
A {\it topological poset} $({\mathcal X},\leq ,{\mathcal T})$ is a poset
$({\mathcal X},\leq )$ equipped with a Hausdorff topology $\mathcal
T$ on the set ${\mathcal X}$ such that the order relation $P \coloneqq \{(x,y)\in
{\mathcal X}\times {\mathcal X}\mid x\leq y\}$ is a closed subspace of
${\mathcal X}\times {\mathcal X}$. This topological poset is said to be
{\it mirrored}, and is called an $M${\it-poset}, if it comes equipped with
a poset map $\mu \colon {\mathcal X}\to {\mathcal R}$, where ${\mathcal R}$
is a finite poset, satisfying
\begin {itemize}
\item $x<y$ implies $\mu (x)<\mu (y)$,
\item for all $r\in {\mathcal R}$, $X_{r} := \mu^{-1}(r)$ is a non-empty closed
subset of ${\mathcal X}$.
\end{itemize}
The map $\mu$ is a {\it mirror} of $\mathcal X$.
\vskip 5pt
It follows that
${\mathcal X}$ is the topological sum $\coprod _{r\in {\mathcal R}}X_r$.
\vskip 5pt
An example is the real {\it Grassmann Poset} ${\mathcal G}_n$; here, ${\mathcal X}$ is the set of all {\it proper} linear subspaces of ${\mathbb R}^{n+1}$ of positive dimension, partially ordered by inclusion, ${\mathcal R}$ is the set $[n]$ of positive integers $\leq n$ with the natural ordering, and $\mu$ takes the $k$-dimensional subspaces to the integer $k$.

\begin{notation} For a subset $A$ of ${\mathcal X}$ define
$${\upset A} \coloneqq \{x\in {\mathcal X}\mid a\leq x \text{ for some } a\in A\}.$$
\end{notation}

\noindent If ${\mathcal A}$ is a family of subsets, $\upset {\mathcal
A} \coloneqq \{\upset A\mid A\in \mathcal A\}$. For $a\in A,$
we use the abbreviation $\upset a \coloneqq \upset \{a\}$.
For $A\subseteq X_r$ and $s>r$, we use the abbreviation $A^{(s)} \coloneqq (\upset A)\bigcap X_s$.
\vskip 5pt

Each $X_r$ inherits a topology from ${\mathcal X}$. We choose a basis ${\mathcal U}_{r}$
for the topology of $X_r$.

\begin{opp} We will assume from now on that our $M$-posets have the
property that when $U$ is open in $X_r$ then $\upset U$ is open in
$\mathcal X$; equivalently, for each $s>r$, the
set $U^{(s)}$ is open in $X_s$.
\end{opp}

The {\it Up topology} on ${\mathcal X}$ has  basis $\bigcup _{r}\{\upset {\mathcal U}_{r}\mid r\in {\mathcal R}\}$. The Openness Property implies that this is a basis for a topology.
\vskip 5pt
It is important to distinguish between the Original topology on ${\mathcal X}$ (which is Hausdorff) and the Up topology on ${\mathcal X}$ (which is $T_0$ but, in general, not $T_1$.) The $M$-posets discussed in this paper are
considered to have both topologies, the ``Up'' being derived from the ``Original''. The Openness Assumption ensures that the Up topology agrees with the ordinary topology on each $X_r$; i.e. the inclusion of $X_r$ into $\mathcal X$ with the Up topology is a topological embedding.

\begin{example} Let ${\mathcal X}$ be the real Grassmann Poset ${\mathcal G}_3$
Here
$X_{1}$ is the collection of lines in ${\mathbb R}^3$ containing $0$ and
$X_{2}$ is the collection of planes in ${\mathbb R}^3$ containing $0$,
partially ordered by inclusion. Separately, the $X_r$ have their Original (usual)
topologies. In the Up topology a basic open neighborhood of a line consists
of an open set of nearby lines together with all planes containing those
lines. A basic open neighborhood of a plane in the Up topology is just a neighborhood of that
plane in the ordinary sense.
\end{example}

\vskip 5pt

\subsection {The Order Complex and the Comparison Map.}\label{order} In \cite{Zi}, a simplicial space $\Delta ({\mathcal X})$, called the {\it order complex}\footnote{Strictly speaking, this is not a complex, but rather the geometric realization of a simplicial topological space.}, is associated with each topological poset ${\mathcal X}$.
For $M$-posets where the spaces $X_r$ are locally compact polyhedra (as will always be the case in this paper) there is a rather simple definition as follows:
\vskip 5pt
Each element $z$ of the topological join ${\Asterisk_{r\in {\mathcal R}}}X_r$ has the form
$z={\sum}t _{i}x_{i}$ where $x_i\in X_i, \\
t _{i}\geq 0, \text{ and } \sum t_i=1$. Define
the {\it support} of $z$ to be
$\text{supp}(z) \coloneqq \{i\in {\mathcal R}\mid t _{i}>0\}$. Then
$$\Delta ({\mathcal X}) \coloneqq
\{z\in {\Asterisk_{r\in {\mathcal R}}}X_r\mid\text{supp}(z)\text{ is
a chain in }{\mathcal R}, \text{ and if }i<j \text{ in supp}(z) \text{ then }
x_{i}<x_{j} \text{ in
}{\mathcal X}\}.$$

\noindent We will sometimes identify $X_r$ with the subspace of $\Delta ({\mathcal X})$ whose support is the singleton $\{r\}$.

\begin{example} Vassiliev \cite{V1}, \cite{V2} proved that for $\mathbb K$ the reals, the
complex numbers, or the quaternions, and
${\mathcal G}_n(\mathbb K)$ the Grassmann Poset of proper non-zero linear
subspaces of $K^{n+1}$, the corresponding order complex,
$\Delta ({\mathcal G}_{n})$, is homeomorphic to the sphere $S^m$ where $m={n+1 \choose 2}d+n-1$, $d$ being the dimension of $K$ over $\mathbb R$.
\end{example}

The {\it Comparison Map} $f \colon \Delta ({\mathcal X})\to {\mathcal X}$ is defined by $f(z)=\text{max}\{x_{i}\mid i\in \text{ supp}(z)\}$. It is not hard to prove that $f$ is continuous\footnote{The word ``map'' is used in this paper to mean that the given function is continuous.} with respect to the Up
topology on ${\mathcal X}$.
\vskip 5pt
Our aim is to prove that under reasonable topological and
geometric assumptions, and considering $\mathcal X$ with the Up topology,
the Comparison Map $f$ is a weak homotopy equivalence. This will be done
by satisfying the hypotheses of the following theorem of McCord \cite{Mc1}, reproved independently\footnote{May's paper appears to date from the late 1970's though the publication year is 2006.} by May \cite{M}.

\begin{thm}\label{mccord} Let $g \colon C\to D$ be a map between topological spaces, and let
${\mathcal B}$ be a basis for the open subsets of $D$. If for all $B\in {\mathcal B}$ the restriction $g| \colon g^{-1}(B)\to B$ is a weak homotopy
equivalence, then $g$ is a weak homotopy equivalence.
\end{thm}
\vskip 5pt
\subsection{Polyhedral $M$-posets}
An $M$-poset ${\mathcal X}$ is {\it polyhedral} if each $X_i$ is a locally compact polyhedron\footnote{In Section \ref{alpha} we go into detail about our use
of polyhedral language.}, and $\mathcal X$ has the {\it Polyhedral Diagonal Property}, namely: the order relation $P$ is a (closed)  subpolyhedron of ${\mathcal X}\times {\mathcal X}$; here $\mathcal X$ has the ordinary topology.
\vskip 5pt
\begin{Rem}\label{polyprop} Checking this property is usually done as follows. There is
an $M$-poset $\mathcal X$ which obviously has the property, and one is interested in a subset $\mathcal Y$ such that  each $Y_i$ is a closed polyhedral subset of $X_i$. Then
the order relation for $\mathcal Y$ is the polyhedron
$P\bigcap {\mathcal Y}\times {\mathcal Y}$, so $\mathcal Y$ is a polyhedral $M$-poset.
\end{Rem}
\vskip 5pt
\subsection{Convenient Choice of Bases} When the space $X_r$ is
polyhedral each point has a basic
system of neighborhoods which are piecewise linear (PL) cones on their frontiers. We may
assume that the basis ${\mathcal U}_r$ consists of such neighborhoods of
points. In particular, each such set is contractible and its closure is
a compact contractible subpolyhedron of $X_r$. {\it Our basis ${\mathcal
U}=\bigcup _{r}\{\upset {\mathcal U}_{r}\}$ for the topology of $\mathcal X$
will always be understood to consist of such sets.}

\vskip 5pt
\subsection{Geometric Posets and Main Theorem}\label{geometric}
A compact subset $C$ of a polyhedron $D$ has {\it trivial shape} if $C$
can be contracted to a point in any of its neighborhoods\footnote{This
is an intrinsic property of $C$, independent of $D$. Note that when $C$
is contractible then $C$ has trivial shape, but trivial shape is more
general: for example, the Topologist's Sine Curve in ${\mathbb R}^2$ has trivial
shape but has two path components.} in $D$.
\vskip 5pt
\begin{notation} When $(M,d)$ is a metric space, the space of non-empty compact
subsets of $M$ with the Hausdorff metric is denoted by $cM$.
\end{notation}
\vskip 5pt
A polyhedral $M$-poset $({\mathcal X},\leq ,{\mathcal T})$ is {\it geometric} if it has the following
\hfil\break {\bf Additional Properties:}
\vskip 5pt
\begin{enumerate}[(1)]
\item [A1]: When $r<s$ and $x\in X_r$, the set $x^{(s)}$ is compact and non-empty.
\item [A2]: When $r<s$, the map $\pi _{r,s} \colon X_{r}\to cX_{s}$ defined by
$x\mapsto x^{(s)}$ is continuous.
\item [A3]: When  $U\in {\mathcal U}_r$, $s>r$, and $y \in \overline
U^{(s)}$, then $\{x\in {\overline U}\mid x<y\}$ has trivial
shape\footnote{This implies that every $y$ has a predecessor $x$.}.
\end{enumerate}
\vskip 5pt
Our goal in this paper is to prove the following theorem:

\begin{thm}\label{main} {\bf (Comparison Theorem)} When the $M$-poset ${\mathcal X}$ is geometric and carries the Up topology,  the map
$f \colon \Delta ({\mathcal X})\to {\mathcal X}$ is a weak homotopy equivalence.
\end{thm}
\vskip 5pt
Our approach will be to prove that for $U\in {\mathcal U}_r$ both $\upset
U$ and $f^{-1}(\upset U)$ are weakly contractible\footnote{i.e. all
homotopy groups are trivial.}, and then to apply
Theorem \ref{mccord}.

\begin{rem} The discrete case of this theorem is due to McCord \cite{Mc2}. See Section \ref{discrete} for a discussion.
\end{rem}

\begin{example} The Grassmann Posets ${\mathcal G}_{n}(\mathbb K)$ are geometric, so Theorem \ref{main} and Vassiliev's result gives us the singular homology of ${\mathcal G}_{n}(\mathbb K)$ with the Up topology.
\end{example}

\section{Connection with Matroids}
\vskip 5pt
While this paper is about topological $M$-posets in general, there is a particular class of such posets which motivates us --- a class associated with tropical phased matroids\footnote{This section can be skipped by readers only interested in the proof of Theorem \ref{main}.}.
\vskip 5pt
\subsection{Tropical phase hyperfield}
Let $T\Phi$ denote the poset $\{0\} \cup S^{1} \subseteq \mathbb{C}$, where the partial ordering is fully defined by $0<x$ for all $x\in S^1$. This is an $M$-poset over $\{0,1\}$, where the mirror $\mu$ takes $0$ to $0$ and all of $S^1$ to $1$. The Openness Property holds, and with the Up topology this is easily seen to be a geometric poset.
\vskip 5pt
$T\Phi $ also carries the structure of a hyperfield. A {\it hyperfield} is similar in definition to a field (associative, commutative etc.) except that the addition operation is allowed to be multivalued. In the case of $T\Phi$, multiplication is inherited from the complex plane and the sum $\boxplus$ is defined by
\begin{itemize}
\item $x \boxplus 0=\{x\}$,
\item $x \boxplus -x=\{0\} \cup S^1$ whenever $x \neq 0$,
\item $x \boxplus y$ is the smallest closed arc in $S^1$ joining the points $x$ and $y$
when $y \neq -x$,
\item For subsets $A$ and $B$, $A\boxplus B=\bigcup _{a\in A, b\in B}a\boxplus b$.
\end{itemize}
$T\Phi $ is the {\it tropical phase hyperfield} introduced by Viro in \cite{Viro1}\footnote{Not to be
confused with the {\it phase hyperfield} where $x\boxplus -x=\{0,x,-x\}$
and ``closed arc" is replaced by``open arc".}.
\vskip 5pt
\begin{rem} The hyperfield $T\Phi$ is the complex analog of the ``sign
hyperfield'' ${\mathbb S}:=T\Phi \bigcap {\mathbb R}$. Just as $T\Phi$
will lead us to a discussion of tropical phased matroids, $\mathbb S$
would lead to an analogous discussion of oriented matroids. However,
unlike $T\Phi$, the original topology on $\mathbb S$ is discrete, so the
analog of the topological posets discussed here would be discrete posets, and everything
would be much simpler.
\end{rem}

\subsection{Relevant $M$-posets}
The product $T\Phi ^{n}$ is an $M$-poset, where the comparison
$<$ is defined coordinate-wise. This has a minimal element
$\mathbf 0$ and is therefore contractible. More interesting
is the subset $T\Phi ^{n}-\{\mathbf 0\}$. The mirror map
$\mu :T\Phi ^{n}-\{\mathbf 0\}\to [n]$ is defined by
$\mu(\mathbf{v})=|\text{supp}(\mathbf{v})|$. This is given the Up
topology. It is polyhedral because the pre-image of each $i$ is
homeomorphic to a disjoint union of tori. That it has the Polyhedral Diagonal Property is then clear.
\vskip 5pt
\begin{prop}\label{product}
This polyhedral $M$-poset is geometric.
\end{prop}
\begin{proof}
The Additional Properties A1 and A2 are clear. With
${\mathcal X}:=T\Phi ^{n}-\{\mathbf 0\}$, recall that Property A3 refers to our convenient choice of basis. In $T\Phi $ this consists of the interiors of closed intervals in $S^{1}$, together with the set $T\Phi $ itself, and in $T\Phi ^{n}-\{\mathbf 0\}$ the basis consists of products of those sets. Let $i<j$ and let $U$ be a basic open set in $X_i$. If
${\mathbf y}\in {\overline U}^{j}$, the set $\{{\mathbf x}\in {\overline U}\mid
{\mathbf x}<{\mathbf y}\}$ is a singleton, obtainable by selecting $i$ entries in the vector ${\mathbf  y}$.
\end{proof}
\vskip 5pt
For an $n$-vector $\mathbf{v}=(v_1,\ldots,v_n) \in T\Phi^{n}-\{\mathbf 0\}$, we define
$$\mathbf{v}^\perp\coloneq\{(x_1,\ldots,x_n) \in T\Phi^{n}-\{\mathbf 0\}\mid 0 \in v_{1}x_1
\boxplus \ldots \boxplus v_{n}x_n\}.$$
\noindent
More generally, given $m$ such $n$-vectors $\mathbf{v}_i \in T\Phi^{n}-\{\mathbf 0\}$, define
$$\{\mathbf{v}_{1}, ..., \mathbf{v}_{m}\}^\perp := \bigcap _{i} \mathbf{v}_{i}^{\perp}.$$
\vskip 5pt
Since $\{\mathbf{v}_{1}, ..., \mathbf{v}_{m}\}^\perp $ is a subset of the $M$-poset
$T\Phi ^{n}-\{\mathbf 0\}$, it is also an $M$-poset.
\vskip 5pt
\begin{thm}\label{perp} The $M$-poset, $\{\mathbf{v}_{1}, ..., \mathbf{v}_{m}\}^\perp$ is geometric.
\end{thm}
\begin{proof}
We first prove polyhedrality. It is enough to prove it for $\mathbf{v}^\perp$. We work by induction on $n$. This is trivial when $n=1$ and is assumed true for vectors of length $<n$. We may assume each $v_{i}x_{i}$ is non-zero, and that the $v_{i}x_{i}$ are distinct; otherwise the claim would be covered by the induction hypothesis. Then $0 \in v_{1}x_{1}\boxplus \ldots \boxplus v_{n}x_{n}$ if and only if either (i) some entry is the negative of another, or (ii) no  open semicircle contains all of the entries. The set of $n$-tuples $(v_{1}x_{1},\ldots ,v_{n}x_{n})$ satisfying (i) is a union of $\frac{n(n-1)}{2}$ tori of dimension $n-1$. The set satisfying (ii) is also easily shown to be polyhedral. By Remark \ref{polyprop} it is a polyhedral $M$ poset. The Additional Properties A1 through A3 hold for the same reasons as in the proof of Proposition \ref{product}.
\end{proof}
\vskip 5pt
\begin{rem} In \cite{Al21}, the first-named author explores the
conditions under which the order complex $\Delta(\{{\mathbf v_1},\dots,{\mathbf v_m}\}^\perp )$
is homeomorphic to the sphere $S^{2(n-m)-1}$. When this holds, Theorem
1.3 implies that, the singular homology of $\{\mathbf{v}_{1}, ..., \mathbf{v}_{m}\}^\perp $
with the Up topology, is that of $S^{2(n-m)-1}$.
\end{rem}
\vskip 5pt
In the case of just one vector ${\mathbf v}$, there is a unique rank
$(n-1)$ ``tropical phased matroid'' whose $T\Phi $ circuit
set is\footnote{Here and below, $T\Phi -\{0\}$ is usefully shortened
to $S^1$.} $S^{1}{\mathbf v}$ and whose covector set is ${\mathbf
{v}}^\perp $. Similarly, a unique tropical phased matroid
is defined by $\{\mathbf{v}_{1},\ldots,\mathbf{v}_{m}\}$ when $m>1$,
but then $S^{1}\mathbf v_{1}\bigcup \ldots \bigcup S^{1}\mathbf v_{m}$
must satisfy certain``circuit conditions''. The rank in that case may
not be $(n-1)$. See \cite{BB1}.

\subsection{Strong matroids over hyperfields}\label{BB}
The only hyperfield occurring in this paper is $T\Phi$. The general
definition of a hyperfield can be found, for example, in \cite{BB1} or \cite{BB2}. To
simplify notation in this subsection it is convenient to state things
in terms of an arbitrary hyperfield $F$, keeping in mind that, for us,
$F$ will
 always be $T\Phi$.
\vskip 5pt
Given a non-empty finite set $E$, a hyperfield $F$, and a positive integer $r$, a {\it strong Grassmann-Pl{\"u}cker function of rank} $r$ on $E$ with coefficients in $F$ is a function $\varphi :E^{r}\to F$ such that
\begin{enumerate}[(1)]
\item $\varphi $ is not identically zero;
\item $\varphi $ is alternating, i.e. $\varphi (x_{1},\dots ,{x}_{i},\dots ,x_{j},\dots x_{r})=-\varphi (x_{1},\dots ,{x}_{j},\dots, x_{i},\dots x_{r})$;
\item $\varphi $ satisfies the strong Grassman-Pl{\"u}cker relations: When $\{x_1,\dots,x_{r+1}\}\subseteq E$ and $\{y_1,\dots,y_{r-1}\}\subseteq E$ then
$$0\in \boxplus _{k=1}^{r+1}(-1)^{k}\varphi (x_{1},\dots ,{\hat x}_{k},\dots x_{r+1}).
\varphi (x_{k},y_{1},\dots ,y_{r-1}).$$
\end{enumerate}
\vskip 5pt
Two such functions $\varphi _1$ and $\varphi _2$ are {\it equivalent} if
$\varphi _1=t.\varphi _2$ where $t\in F$, i.e. one is a scalar multiple
of the other.
\vskip 5pt
Strong $F$-matroids are defined in \cite{BB1}. For our purposes we will instead use the following Theorem 3.13 of \cite{BB1} as our {\it definition} of ``weak $F$-matroid".
\vskip 5pt
\begin{thm}\label{strong} There is a natural bijection between the set of equivalence classes of strong  Grassman-Pl{\"u}cker functions of rank $r$ on $E$ with coefficients in $F$ and strong  $F$-matroids of rank $r$ on $E$.
\end{thm}

\subsection{Srong tropical phased Grassmannians}\label{circuit}
We apply the previous subsection with $E=[n]$ and $F=T\Phi $. A map
$\varphi :E^{r}\to T\Phi $ assigns an element $v(i_{1},\ldots.i_{r})\in T\Phi$ to each $r$-tuple $(i_{1},\ldots.i_{r})$ of elements of $[n]$. So $\varphi $ is encoded by an element of $T\Phi ^{n^{r}}$. The space of all maps $\varphi :E^{r}\to T\Phi $ is thus homeomorphic to the product space $T\Phi ^{n^{r}}$. We wish to restrict attention to $T\Phi ^{n^{r}}-{\mathbf 0}$, the space of non-zero $r$-tuples. This is an $M$-poset over $[r]$ whose mirror map takes each $r$-tuple to the size of its support.  As before (compare Proposition \ref{product}) the poset $T\Phi ^{n^{r}}-\{\mathbf 0\}$ with the Up topology is geometric. The quotient space
$(T\Phi ^{n^{r}}-\{\mathbf 0\})/S^{1}$ can be identified with the subspace of $T\Phi ^{n^{r}}-\{\mathbf 0\}$ consisting of normalized $r$-tuples, where the $r$th entry is either $1$ or $0$. This is a geometric $M$-poset whose mirror map is denoted by
$\mu :T\Phi ^{n^{r}}-\{\mathbf 0\}\to [r-1]$. Each $\mu ^{-1}(i) $ is a disjoint union of tori.
\vskip 5pt
The {\it strong Grassmannian}, $\text{Gr}_{s}(r,T\Phi ^{n})$, is defined in the literature\footnote{See \cite{AD} for more details.} to be the topological poset of all strong tropical phased matroids of rank $r$ on the set $[n]$ with
coefficients in $T\Phi $. In view of the previous paragraph, Theorem \ref{strong} identifies this set with the subset of $T\Phi ^{n^{r}}-\{\mathbf 0\}$ consisting of normalized non-zero anti-symmetric functions $\varphi $ satisfying the strong Grassman-Pl{\"u}cker relations.
\vskip 5pt
The set of $r$-tuples of elements of $[n]$ is $[n]^r$. A {\it transposition} of an $r$-tuple is an $r$-tuple obtained by interchanging two entries. In particular, every $r$-tuple containing the same element in two different positions is a transposition of itself. We need a combinatorial lemma:
\vskip 5pt
\begin{lemma}\label{combo} There is a subset $T$ of $[n]^r$ with the following properties.
\begin{itemize}
\item No $r$-tuple having two equal entries is in $T$;
\item Every $r$-tuple whose entries are all different is obtainable from an element of $T$ by a transposition;
\item No transposition of an element of $T$ is in $T$.
\end{itemize}
\end{lemma}
\begin{proof}
Define $S_{1}:=\{s\in [n]^r\mid \text{ all entries in } s\text{ are
different}\}$.  Choose a total ordering of $S_1$. Let $m_{1}$ be the least
element of $S_1$.
Define $$S_{2}:=\{s\in S_{1}\mid s>m_{1}\text{ and } s \text{ is not obtainable
from }m_{1}\text{ by a transposition}\}.$$
If
$S_2$ is non-empty let $m_{2}$ denote its minimal element; in particular,
$m_{2}>m_{1}$.
Let $$S_{3}:=\{s\in S_{2}\mid s>m_{2}\text{ and } s \text{ is not obtainable
from }m_{2} \text{ by a transposition}\}.$$
Proceed in this
way until reaching $d$ such that $S_{d+1}$ is empty. The resulting set
$T:=\{m_{1},\ldots ,m_{d}\}\subseteq [n]^r$ has the required properties,
and $|T|=d$.
\end{proof}
\begin{thm}\label{grass}
As a topological poset $\text{Gr}_{s}(r,T\Phi ^{n})$ is geometric.
\end{thm}
\begin{proof}
It follows from Lemma \ref{combo} that the space of alternating
functions $\varphi $ is homeomorphic to the product space $T\Phi
^{d}$. This is because once $d$ such functions are chosen using the Lemma
then the others are determined. Thus the space of non-zero antisymmetric
functions $[n]^{r}\to T\Phi$ is polyhedral. The subspace consisting of those
functions which satisfy the strong Grassman-Pl{\"u}cker
relations is therefore polyhedral for the reasons explained in the proof
of Theorem \ref{perp}; see in particular Remark \ref{polyprop}. The Additional Properties A1 through A3 hold for the same reasons as in the proof of Proposition \ref{product}.
\end{proof}
\vskip 5pt
The ordinary complex Grassmannian $\text{Gr}(r,{\mathbb C}^{n})$ is the
space of $r$-dimensional linear subspaces of ${\mathbb C}^{n}$; here $1\leq r\leq n-1$. But it can also be viewed in the above terms, since every field is a hyperfield (in which case $a+b=\{a+b\}$). Thus it can be identified with the space of normalized strong Grassmann-Pl{\"u}cker functions of rank $r$ on $[n]^{r}$ with coefficients in $\mathbb C$.
\vskip 5pt
The canonical map ${\mathbb C}\to T\Phi$ sending non-zero $z$ to
$\frac{z}{|z|}$ and $0$ to $0$ is continuous with respect to the Up
topology. It induces the map $\rho _{s}:\text{Gr}(r, {\mathbb C}^{n})\to
\text{Gr}_{s}(r, T\Phi^{n})$, which is also continuous.
\vskip 5pt
The following is the tropical phased analog of a well-known conjecture for oriented matroids:
\vskip 5pt
{\bf Question:} Is the map $\rho _{s}:\text{\rm Gr}(r, {\mathbb C}^{n})\to \text{\rm Gr}_{s}(r, T\Phi^{n})$ a weak homotopy equivalence?
\vskip 5pt
\begin{rems}
\begin{enumerate}[(i)]
\item There is a homotopy commutative diagram
$$
\xymatrix{
&\Delta (\text{Gr}_{s}(r,T\Phi^{n}))\ar[d]^{f}\\
\text{Gr}(r,{\mathbb C}^{n})\ar[r]_{\rho}\ar[ur]^{\Delta (\rho)}&\text{Gr}_{s}(r,T\Phi^{n})
}
$$
and, in view of Theorem \ref{main}, the map $\rho $ is a weak homotopy equivalence if and only if $\Delta (\rho)$ is.
\item The analogous ``MacPhersonian Conjecture" for oriented matroids
(still open having been incorrectly dealt with in \cite{biss}) says
that the analogous ``real" map $\Delta (\rho):\text{Gr}(r, {\mathbb
R}^{n})\to \Delta (\text{Gr}(r, {\mathbb S}^{n}))$ is a weak homtopy
equivalence. Since $\text{Gr}(r,{\mathbb S}^{n}))$ is a finite
discrete $M$-poset, the relevant analog of Theorem \ref{main} was proved long
ago by McCord (see Section \ref{discrete}), and it was already pointed
out in \cite{AD} that the MacPhersonian Conjecture is the same whether worded for $\rho $
or for $\Delta (\rho)$.
\end{enumerate}
\end{rems}
\vskip 5pt
{\bf Summary:} The point of this section is to bring out how interest
in Theorem \ref{main} has arisen in connection with matroids. Theorems
\ref{perp} and \ref{grass} are instances where one can travel back and
forth between the order complex and the actual topological poset equipped with the Up topology.
\vskip 5pt
The rest of this paper is concerned with the proof of Theorem \ref{main}.
\vskip 5pt
\section{Preliminaries}

\subsection{Polyhedra.}\label{alpha} This is for reference. We need to be precise about our polyhedral terminology.
\vskip 5pt
A {\it rectilinear simplicial complex} consists of a locally finite, finite-dimensional, abstract simplicial complex $K$, whose geometric realization is denoted by $|K|$, and  a closed proper  topological embedding of $|K|$ in a euclidean space $E$ which restricts to an affine isometric embedding of each simplex of $|K|$. The choice of $E$ and of the embedding will be suppressed, and from now on we will denote its image simply by $K$, omitting the vertical bars.
\vskip 5pt
Each simplex of $K$ has a euclidean metric. The ``length metric'' on $K$ is the metric which  measures the distance between points $p$ and $q$ as the {\it inf} of lengths of piecewise linear paths joining $p$ to $q$, where each linear part of the path lies in (and is measured in) a simplex of $K$. The corresponding metric topology on $K$ is the same as both the weak topology and the topology inherited from the euclidean space $E$.
\vskip 5pt
A locally compact space $A$ is a {\it polyhedron} if it is equipped
with a piecewise linear (PL) structure. This means that the space $A$
has been identified, via a homeomorphism chosen once and for all,
with a rectilinear simplicial complex $K$, and the PL {\it structure}
consists of all rectilinear simplicial complexes $L$ occupying the same
space as $K$, such that $L$ and $K$ have a common rectilinear simplicial
subdivision\footnote{References for polyhedra are \cite{RS}, \cite{Hd}
and Section 3.1 of \cite{Sp}. Here, we use nothing of that subject
beyond basic definitions and elementary properties.}. We say that any
such complex $L$ {\it triangulates} $A$.
\vskip 5pt
A {\it subpolyhedron} of $A$ is a closed subset $B$ such that some triangulation of $A$ in the PL structure has a subcomplex which triangulates $B$. We will often describe a space $A$ as {\it polyhedral} when we mean that $A$ is a polyhedron in the above sense; and when $B$ is a subpolyhedron of $A$ we simply say that $B$ is ``polyhedral''.
\vskip 5pt

As an extension of this, we may consider an open subset $V$ of the
polyhedron $A$. It is well-known that $V$ has a polyhedral structure
{\it compatible with} that of $A$; this means: $V$ can be triangulated
in such a way that each finite subcomplex is a subpolyhedron of $A$,
and the diameters of simplexes approach zero as the simplexes approach
the frontier of $V$.

\vskip 5pt
The polyhedron $A$ has many length metrics, one for each triangulation $L$. In particular, small metric balls in polyhedra are cones on their frontiers.  Once a length metric is chosen then, for small enough $\eta >0$, two maps from a compact domain into $A$ which are distant at most $\eta$ apart pointwise are $\eta $-homotopic\footnote{Maps $a,
b \colon A\to B$ are $\eta${\it -homotopic} if there is a homotopy between $a$
and $b$ such that for each point $x\in A$ the diameter of the image of
$x\times I$ is $<\eta$.}.

\begin{rems}
\begin{enumerate}[1.]
\item The class of polyhedra is closed under coproducts and finite products.
\item If $B$ is a compact polyhedral subset of a finite product $A$ of polyhedra, then projections of $B$ to sub-products of $A$ are also compact polyhedral.
\item The join of two compact polyhedra is a compact polyhedron.
\item Smooth manifolds are homeomorphic to polyhedra.
\item The topological join of two locally compact spaces need not be
locally compact, so if they are polyhedra, their join does not satisfy
our definition of ``polyhedron''. However, the join of two abstract
simplicial complexes has an obvious meaning even when they are not locally
finite. Therefore the topological join of two (hence of finitely many)
polyhedra can be triangulated by a simplicial complex which might not
be locally finite. Occasionally, we will allow ourselves to extend the word
``polyhedral'' to include this case.
\end{enumerate}
\end{rems}
\vskip 5pt

\subsection{Some Consequences of the Additional Properties}
\begin{prop}\label{upcompact}When $r<s$ and $A$ is a compact subset of $X_r$ then $A^{(s)}$ is a compact subset of $X_s$.
\end{prop}
For this we need notation and a lemma.

\begin{notation} When $(M,d)$ is a metric space and $S\subseteq cM$, we define
$\bigcup(S) \coloneqq \{m \in M\mid m \in s\text{ for some }s\in S\}$, i.e. $\bigcup(S)$ is the union of the sets in $S$.
\end{notation}

\begin{lemma}\label{hyper} Let $C$ be a compact subset of $c{\mathbb R}^{n}$. Then $\bigcup(C)$ is a compact subset of $\mathbb{R}^n$.
\end{lemma}
\begin{proof} $C$ is totally bounded in the Hausdorff metric, so $\bigcup(C)$ is a bounded subset of ${\mathbb R}^{n}$. If $x$ is a limit point of $\bigcup(C)$ there exists $(x_{n})$, a sequence in $\bigcup(C)$, converging to $x$. Let $x_{n}\in c_{n}$ where $c_{n}\in C$. Without loss of generality, assume
$\{c_{n}\}$ converges to $c\in C$ in the Hausdorff metric. So $x$ must lie in $c$.
Thus $\bigcup(C)$
is a closed and bounded subset of ${\mathbb R}^{n}$.
\end{proof}

\begin{proof}[Proof of Proposition \ref{upcompact}] By the definition of a geometric poset the set $\pi _{r,s}(A)$ is a compact subset of $cX_s$. The polyhedron $X_s$ can be regarded as a subset of a euclidean space, so Lemma \ref{hyper} implies $A^{(s)}=\bigcup(\pi _{r,s}(A))$ is compact.
\end{proof}
\vskip 5pt
\begin{prop}\label{poly}
Let $A$ be a compact subset of $X_r$ and let $r\leq s$. If $A$ is polyhedral then $A^{(s)}$ is polyhedral.
\end{prop}
\begin{proof}
Define $B \coloneqq (A\times X_{s})\bigcap P$. Then $B =\{(a,x)\mid
a<x\}\subseteq A\times A^{(s)}$ is polyhedral. The $X_s$ projection
of $B$ is $\{x\mid a<x \text{ for some }a\in A\}$. This is $A^{(s)}$,
and is compact by Proposition \ref{upcompact}. There are compact polyhedra $R\subseteq
X_r$ and $S\subseteq X_s$ such that
$A\times A^{(s)}\subseteq R\times S$. Since $B$ is polyhedral and is
a subset of $R\times S$, $B$ is compact polyhedral. Thus the projection
of $B$, namely $A^{(s)}$, is polyhedral.
\end{proof}
\vskip 5pt

\subsection {Tools from geometric topology}\label{tools}
Here we state four theorems, well-known in geometric topology but perhaps less well-known in other mathematical communities.
\vskip 5pt
Recall that a space $Z$ is {\it locally $n$-connected} if for each $z\in Z$ and each neighborhood $U$ of $z$ there is a neighborhood $V$ of $z$ such that, for all $k\leq n$, every map from the $k$-sphere $S^k$ into $V$ extends to a map of the $(k+1)$-ball $B^{k+1}$ into $U$. For us, the important point is that polyhedra are locally $n$-connected for all $n$.
\vskip 5pt
\begin{thm}\label{hu1} Let $Z$ be a locally $n$-connected metrizable space. Every open cover
$\mathcal A$ of $Z$ has an open refinement $\mathcal B$ such that whenever $K$ is a  polyhedron of dimension $\leq n$, any two maps $g_{1}, g_{2}:K\to Z$ which agree on the closed subset $C\subseteq K$ and are $\mathcal B$-close, are $\mathcal A$-homotopic rel $C$.
\end{thm}
\begin{proof} This is Theorem V.5.1 of \cite{Hu}. The references to $C$ are not in the statement, but they follow from the proof.
\end{proof}
\vskip 5pt
When $K$ is a simplicial complex and $L$ is a subcomplex containing all the vertices of $K$, a {\it partial realization} of $K$ in a space $Z$ relative to a cover $\mathcal A$ of $Z$ is a
map $\psi :L\to Z$ such that, for every simplex $\sigma $ of $K$,  there is a set $A\in {\mathcal A}$ with $\psi (\sigma\bigcap L)\subseteq A$. When $L=K$ this is a {\it full realization}.
\vskip 5pt
\begin{thm}\label{hu2} Let $Z$ be a locally $(n-1)$-connected metrizable space,  Every open cover $\mathcal A$ of $Z$ has an open refinement $\mathcal B$ such that every partial realization $j_{0}:L\to Z$ relative to $\mathcal B$ extends to a full realization $j:K\to Z$ relative to $\mathcal A$.
\end{thm}
\begin{proof}
This is Theorem V.4.1 of \cite{Hu}.
\end{proof}
\vskip 5pt
Recall that a map is {\it proper} if the pre-image of every compact subset of its codomain is compact. A proper map $q:A\to B$ between locally compact polyhedra is {\it cell-like} if
$q^{-1}(b)$ has trivial shape\footnote{See Section \ref{geometric}.} for every $b\in B$.
\vskip 5pt
\begin{thm}\label{lacher} Let $A$ and $B$ be locally compact metric
spaces, $K$ a finite-dimensional, locally finite simplicial complex, $L$
a subcomplex of K, $\theta :A\to B$ a (proper) cell-like map, $\varphi :K\to
B$ a proper map, $\psi :L\to A$ a proper map such that $\theta \circ \psi =
\varphi\mid L$, and $\epsilon :B\to (0,\infty)$ a function. Then there
exists a proper map ${\widebar \varphi}:K\to A$ such that
$d(\theta \circ {\widebar \varphi}(x), \varphi (x))\leq \epsilon (\varphi (x))$ for every $x\in K$.
\end{thm}
\begin{proof} This is Lemma 2.3 of \cite{L}.
\end{proof}
\vskip 5pt
\begin{thm}\label{lacher2} A cell-like map between locally compact polyhedra is a proper homotopy equivalence.
\end{thm}
\begin{proof} This is Theorem 1.5 of \cite{L}.
\end{proof}
\vskip 5pt

\section{The special case ${\mathcal R}=[2]$}\label{2case}
\begin{notation} In this section, when $A\subseteq X_{1}$ we write $A'$ rather than $A^{(2)}$.
\end{notation}
\vskip 5pt
Nearly all the ideas needed for our proof of Theorem \ref{main} occur
when dealing with two special cases, ${\mathcal R}=[2]$ and ${\mathcal R}=[3]$, where notation is less cluttered. We treat ${\mathcal R}=[2]$ here, and ${\mathcal R}=[3]$ in the following section.
\vskip 5pt
When ${\mathcal R}=[2]$, the geometric $M$-poset is ${\mathcal X}=X_{1}\coprod X_{2}$; $\mu (X_{1})=1$ and
$\mu(X_{2})=2$.
\vskip 5pt
In this case, the join $X_{1}*X_{2}$ consists of a copy of $X_{1}$, a copy of $X_{2}$, and for
each $x\in X_{1}$ and $y\in
X_{2}$ a linearly parametrized path $\omega$ with $\omega (0)=x$ and $\omega
(1)=y$. We call such a path (or its image) a {\it segment} from $x$
to $y$. The point $\omega (t)$ on a segment from $x$ to $y$ is often
more conveniently denoted by $(1-t )x+ty$ where $0\leq t\leq1$. The
map $q\colon X_{1}\times X_{2}\times I\to X_{1}*X_{2}$ taking $(x,y,t)$ to $(1-t)x+ty$ defines the
quotient topology on $X_{1}*X_{2}$.
\vskip 5pt

\subsection{Weak contractibility of basis elements}\label{2.2}
Basis elements lying in $X_2$ are contractible. Here we prove that when $U$ is a (contractible) basis element in $X_1$ then $\upset U$ is weakly contractible; equivalently, that for each $n$,
$\pi _{n}(\upset U, U)$ is trivial.

\begin{Def} Let $Z$ be a space, and let $A\subseteq X_1$. A map $s' \colon Z\to A'$ {\it covers} a map $s \colon Z\to A$ if $s(z)<s'(z)$ for every $z\in Z$.
\end{Def}
\vskip 5pt
Fundamental to understanding homotopies in this context is the following
\vskip 5pt
\begin{prop}\label{cover}
If $s'$ covers $s$ then, as maps $Z\to \upset A$, $s'$ and $s$ are homotopic.
\end{prop}
\begin{proof}
Define $F \colon Z\times I\to \upset A$ by
\begin{enumerate}[(i)]
\item $F=s'$ on $Z\times [0,1)$;
\item $F=s$ on $Z\times \{1\}$.
\end{enumerate}
We check continuity of $F$. When $W$ is open in $A'$ then
$F^{-1}(W)=s'^{-1}(W)\times [0,1)$ which is open in $Z\times I$. When $V$ is open in $A$ then
\begin{align*}
F^{-1}(V\cup V') & = F^{-1}(V)\cup F^{-1}(V')\\
& = [s^{-1}(V)\times \{1\}]\cup [s'^{-1}(V')\times [0,1)]
\end{align*}
Since $s^{-1}(V)\subseteq s'^{-1}(V')$ we conclude that $F^{-1}(\upset V)$ is open in
$Z\times I$.
\end{proof}
\vskip 5pt
We abbreviate the $n$-ball $B^n$ to $B$, and we consider an arbitrary map
$$g \colon (B,\partial B)\to (\upset U, U).$$
\noindent We write $C \coloneqq g^{-1}U$,
a closed subset of $B$ containing $\partial B$. We want to produce maps $h'$
and $h$ from $B$ to $\upset U$ satisfying:

\begin{enumerate}[(i)]
\item $h'=h=g$ on $C$;
\item $h'$ maps $B-C$ into $U'$;
\item $h$ maps $B$ into $U$;
\item on $B-C$, $h'$ covers $h$;
\item $h'$ is homotopic to $g$ rel $C$.
\end{enumerate}

\noindent Proposition \ref{cover} will then imply that $h'$ is homotopic to $h$ rel $C$
and hence that $g$ represents the trivial element of $\pi _{n}(\upset U, U)$.
\vskip 5pt
We need some preparatory lemmas.
\vskip 5pt
\begin{lemma}\label{relation} If $\{V_m\}_{m=1}^\infty$ is a sequence of neighborhoods of $x\in X_{1}$ such that $\bigcap\limits_{m=1}^\infty V_m=\{x\}$, then
$\bigcap\limits_{m=1}^\infty V_{m}'=x'$.
\end{lemma}
\begin{proof}
It is clear that $x' \subseteq \bigcap\limits_{m=1}^\infty V_{m}'$.
To show containment in the other direction, let $y \in \bigcap\limits_{m=1}^\infty V_{m}'$.
It follows that for each $m$, there exists $x_m \in V_m$ such that $x_{m}<y$ in
$\mathcal{X}$. Since $\bigcap\limits_{m=1}^\infty V_{m}=\{x\}$, $x_m$ converges to
$x$ as $m\to \infty$. Thus, $(x_{m},y)$ converges to $(x,y)$.
Since $\{(x_m,y)\} \subseteq P$ and the order relation $P$ is closed in $\mathcal{X}^2$, $(x,y) \in P$.
\end{proof}
\vskip 5pt
\begin{lemma}\label{yznearx} Let $x\in {\overline U}$. Given $\epsilon
>0$, there exists $\eta (x)>0$ such that whenever\footnote{$B_{\delta}$
denotes the ball of radius $\delta$; $N_{\epsilon}$ denotes the $\epsilon
$-neighborhood.} $y$ and $z$ lie in $B_{\eta}(x)$ then $z'\subseteq
N_{\epsilon}(y')$.
\end{lemma}
\begin{proof} This is an immediate consequence of Property A2.
\end{proof}
\vskip 5pt
\begin{lemma}\label{uly} Given $\epsilon >0$ there exists $\delta >0$
such that whenever $p\in C$ and $q\in B_{\delta}(p)-C$ then $g(q)\in
N_{\epsilon}(g(p)')$.
\end{lemma}
\begin{proof} We write ${\mathcal C} \coloneqq \{\upset B_{\eta (x)}(x)\mid x\in {\overline U}\}$, where
$\eta(x)$ is as in Lemma \ref{yznearx}. Then $g^{-1}({\mathcal C})$ covers the compact set $C$. Let $\delta $ be a Lebesgue number for this cover. For some $x\in {\overline U}$,
$g(B_{\delta}(p))\subseteq \upset B_{\eta (x)}(x)$. This means that
for some $y\in B_{\eta (x)}(x)$, $g(q)\in y'$. By Lemma \ref{yznearx},
$g(q)\in N_{\epsilon}(g(p)')$.
\end{proof}
\vskip 5pt
It turns out to be easier to work with the closure, ${\overline U}$, than with $U$, because ${\overline U}$ is compact. At the end, we will show how to get the desired result for $U$.
\vskip 5pt
Let $Q:=({\overline U}\times {\overline U}')\bigcap P$. By hypothesis, $Q$
is a compact polyhedron. The metric on ${\overline U}\times {\overline
U}'$ is taken to be the sum of the separate metrics on $\overline U$
and ${\overline U}'$. Let ${\mathcal A}_{i}$ be an open cover of $Q$
by sets of diameter $<\frac{1}{i}$, and let ${\mathcal B}_i$ be the
refinement given by Theorem \ref{hu2}. Let $4\lambda _i$ be a Lebesgue
number for ${\mathcal B}_i$.

\vskip 5pt

For each $x\in B-C$ we choose a point $r(x)\in C$ realizing
$nd(x,C)$. Define $k_{0}(x):=gr(x)\in {\overline U}$. Since $g$ is continuous, Lemma
\ref{uly} implies that there exists $\gamma _{i}$, decreasing to $0$ as $i\to \infty$,
such that when $x\in I_{i}:=\{x\in B-C\mid \gamma _{i+1}\leq d(x,C)\leq
\gamma_{i}\}$ then, for some point $k_{0}'(x)$ in the compact non-empty
set $k_{0}(x)'\subseteq {\overline U}'$, $d(k_{0}'(x),g(x))\leq
\lambda _{i}$. Then $k_{0}(x)<k_{0}'(x)$ in the poset. We write
\footnote{It is not claimed that $j_0$ is continuous. We will only be interested in its restriction to the (discrete) $0$-skeleton of a chosen triangulation.}
$j_{0}:=(k_{0},k_{0}'):B-C\to Q$.

\includegraphics[width=11cm]{ag35}

\vskip 5pt
The map $g$ is uniformly continuous on the compact set $I_i$, so, there exists
$\theta _{i}$ such that when $S$ is a subset of $I_i$ of diameter $<\theta _{i}$ then $g(S)$ has diameter $<\lambda _{i}$.
\vskip 5pt
We may assume the sequences $\{\theta _{i}\}$ ais decreasing and converges to $0$.
\vskip 5pt
Assume that each $I_i$ is a
subpolyhedron\footnote{This makes the exposition less cluttered. If $I_i$ is not polyhedral one covers it by a slightly larger polyhedral set. We omit the details.} of $B-C$.  Let $K$ be a triangulation of $B-C$ such that
each simplex lies in some $I_i$, and simplexes in $I_i$ have diameter
$<\theta _i$. When a simplex $\sigma $ lies in $I_i$, the diameter of $k_{0}'({\sigma }^{0})$ is  $<3\lambda _{i}$.
\vskip 10pt
The restriction $g\mid :C\to {\overline U}$ is uniformly continuous, so  there exists
$\rho (\lambda _{i})$ such that when $r(\sigma ^{0})$ has diameter $<\rho (\lambda _{i})$ then $gr(\sigma ^{0})$ has diameter $<\lambda_{i} $. We may assume the sequence
$\{\rho (\lambda _{i})\}$ is decreasing and converging to $0$.
\vskip 5pt
For any integer $\ell$ and simplex $\sigma $ lying in $I_{\ell }$, $r(\sigma ^{0})$ has diameter
$<2\gamma _{\ell }+\theta _{\ell }$. We choose an increasing sequence $\{\ell (i)\}$ such that $2\gamma _{\ell (i)}+\theta _{\ell (i)}<\rho (\lambda _{i})$. Then, for $\sigma $ lying in
$I_{\ell (i)}$, $k_{0}(\sigma ^{0})$ has diameter $<\lambda _{\ell (i)}\leq \lambda_{i}$. Summarizing:
\vskip 5pt
\begin{prop} When $\sigma$ is a simplex of $I_{\ell (i)}$, then
$k_{0}(\sigma ^{0})$ has diameter $<\lambda _{i}$, and $k_{0}'(\sigma
^{0})$ has diameter $<3\lambda _{i}$.\qed
\end{prop}
\vskip 5pt
Let $J_{\ell (i)}:=\bigcup _{\ell (i)\leq m\leq \ell (i+1)} I_{m}$. Then for every simplex
$\sigma $ of $J_{\ell (i)}$ the set $j_{0}(\sigma ^{0})$
has diameter $<4\lambda _i$ and thus lies in an element of the cover
${\mathcal B}_i$ of $Q$. The sequence $\{\ell (i)\}$ starts with $i=0$
where we interpret $\frac{1}{0}=\infty$. The cover ${\mathcal A}_{0}$
is the singleton $\{Q\}$, and ${\mathcal B}_{0}$ has Lebesgue number
$4\lambda _{0}$.
\vskip 5pt
Let $J:=\bigcup _{i\geq 0}J_{\ell (i)}$, regarded as a subcomplex of
$K$. We wish to extend $j_{0}\mid :J^{0}\to Q$ to a map $j:J\to Q$. We do
this first on $J_{\ell (0)}$ where $j_{0}$ maps the $0$-skeleton of
each simplex to a set of diameter $<4\lambda _{0}$. Theorem \ref{hu2}
gives an extension $j$ on $J_{\ell (0)}$. Next, we consider $j_{0}^+$ on
$J_{\ell (1)}$ where $j_{0}^{+}=j_{0}$ on the vertices of $J_{\ell (1)}$
and $j_{0}^{+}$ agrees with the (previously defined) $j$ on $J_{\ell
(0)}\bigcap J_{\ell (1)}$. We subdivide this intersection so that the
$j$-image of each simplex in it has diameter $<\lambda _{1}$, and we
extend this subdivision to  $J_{\ell (0)}\bigcup J_{\ell (1)}$ without
adding further vertices. Then we have a partial realization of
$J_{\ell (1)}$ in $Q$ relative to ${\mathcal B}_{1}$, and Theorem \ref{hu2}
gives an extension of $j$ to all of $J_{\ell (1)}$ extending the map $j$
previously defined on $J_{\ell (0)}$. The $j$-image of each simplex in
$J_{\ell(1)}$ lies in an element of ${\mathcal A}_{1}$.  We continue
in this way, so that on $J_{\ell (i)}-\text {\rm int}J_{\ell(i-1)}$ the
$j$-image of the $0$-skeleton of each simplex has diameter  $<4\lambda
_{i}$, extending using Theorem \ref{hu2}.
\vskip 5pt
Let $N:=J\bigcup C$. This is a neighborhood of $C$ in $B$. The map
$j=(k,k')$; i.e. its components are $k:J\to {\overline U}$ and $k':J\to {\overline U}'$. We extend both of
these maps\footnote{Previously, we defined $k_0$ and $k'_0$
on all of $K$. We now confine the domains of $k_0$ and $k'_0$ to $J$. In
what follows we need to be free to extend $j$, i.e. $(k, k')$, to all
of $K\bigcup C$ in a different way.} (without change of label) to the rest of $N$, agreeing with $g$ on $C$.
\vskip 5pt
\begin{prop}\label{continuous} The extended functions $k:N\to Q$ and
$k':N\to {\overline U}\coprod {\overline U'}$ are continuous.
\end{prop}
\begin{proof}The only issue is continuity at any point $p\in C$. Let ${\upset T}$ be a basic (open) neighborhood of $g(p)$. Since $g$ is continuous $g^{-1}({\uparrow T})$ is an open
neighborhood of $p$. It is enough to find a smaller neighborhood, $W$, of $p$ such that $k'(W)\subseteq T'$ and $k(W)\subseteq T$.
\vskip 5pt
If $\{{\overline V}_{m}\}$ is a basic sequence of compact neighborhoods
of $g(p)$ in $U$, then $\bigcap \{{\overline V}_{m}'\}=g(p)'$
by Lemma \ref{relation}.
By Property A1, $g(p)'$ is a compact non-empty subset of
the open set $T'$, so for sufficiently large $m$ the
compact sets ${\overline V}_{m}'$ and the (closed) frontier of $T '$ are
disjoint. Choose $\epsilon $ so that $2\epsilon =d(g(p)', \text{fr}(T'))$. For this choice, let $\delta$ be as in Lemma \ref{uly}. The following shows that $B_{\delta}(p)$ is the required neighborhood $W$.
\vskip 5pt
When $x$ is a point of $K$ such that $d(x,p)<\delta$ then $d(x,r(x))<\delta$. By Property A1, $g(r(x))'$ is compact and non-empty.
Since $d(x,r(x))<\delta$, $d(g(x),g(r(x))')<\epsilon$,
hence $d(g(x),k'(x))<\epsilon$. And since $d(x,p)<\delta$,
$d(g(x),g(p)')<\epsilon$. Hence $d(k'(x), g(p)')<2\epsilon$. This implies
$k'(x)\in T'$.
\vskip 5pt
Continuity of $k$ at $p$ follows from the fact that $g$ is continuous
at $p$: given $\epsilon >0$ let $\delta >0$ be such that $g$ maps the
$\delta $-neighborhood of $p$ into the $\epsilon $-neighborhood of
$g(p)$ in $U$. If $d(p,x)<\frac{\delta }{2}$ then $d(p,r(x))<\delta $,
so $d(g(p), gr(x))<\epsilon$; i.e. $d(k(p),k(x))<\epsilon$. Thus
$k$ is continuous at $p$.
\end{proof}
\vskip 5pt
The required maps $h$ and $h'$ are defined to agree with $k$ and $k'$ respectively on $N$. Then $h=h'=g$ on $C$, and $h'$ covers $h$ (on $N$).
\vskip 5pt
\begin{rem} Theorem \ref{hu1} implies that, with appropriate care, we may arrange that $h'|N$ and $g|N$ are homotopic rel $C$. The details are left to the reader.
\end{rem}
\vskip 5pt
It remains to extend $h$ and $ h'$ to the rest of $K$ maintaining the properties (i)-(v).
\vskip 5pt
\includegraphics[width=11cm]{ag2}
\vskip 5pt
Recall that $Q:=({\overline U}\times {\overline U}')\bigcap P$. The projection map
$\pi :({\overline U}\times {\overline U}')\to {\overline U}'$ is easily seen to restrict to a proper map $Q\to {\overline U}'$,
and, by Property A3, this map is cell-like.
\vskip 5pt
Let $L$ be the finite simplicial complement of $N$ in $B$; i.e.
$L\bigcup J=K$, and $L$ is disjoint from the interior of the neighborhood $N$.
We write ${\overset \bullet L}=L\bigcap N$, the frontier of $N$ in $B$. We
have a map $(L\times 0)\bigcup ({\overset \bullet L}\times I)\to {\overline U}'$
agreeing with $g$ on $L\times 0$ and being a homotopy between $g|{\overset
\bullet L}$ and $h'|{\overset \bullet L}$ on ${\overset \bullet L}\times
I$. The Homotopy Extension Property gives a map ${\widebar g}:L\to {\overline U}'$
homotopic to $g$ rel $C$ and agreeing with $h'$ on ${\overset \bullet
L}$. Now, $h'|:{\overset \bullet L}\to {\overline U}'$ lifts, through $\pi$, to a map
${\overset \bullet L}\to Q$. Theorem \ref{lacher} then gives an ``almost lift" $j$ of ${\widebar g}$ through $\pi$, continuously extending that map to all of $L$.

$$
\xymatrix{
{\overset \bullet L} \ar[d]_{\text{inclusion}} \ar[r]^{j\mid} &Q\ar[d]^{\pi\mid}\\
L \ar[r]^{\widebar g} \ar[ur]^{j} &{\overline U}'}
$$

Where the domains overlap,
this $j$ agrees with that previously defined on $N$, so, gluing the
two maps together, we have $j:K\to Q$. The components of $j$ are the required maps $h$ and $h'$. They extend the previously defined $h$ and $h'$. By Theorem \ref{lacher}, the component $h' (=\pi \circ j)$ can be as close as we please to $\widebar g$, and hence is homotopic to it. Thus $g$ is homotopic, rel $\partial B$, to a map
whose image misses ${\overline U}'$. In other words:
\begin{thm}\label{2stage}For all $n$, $\pi _{n}(\upset {\overline U}, {\overline U})=0$; i.e. the set $\upset {\overline U}$ is weakly contractible.
\end{thm}
\begin{cor}\label{open} The set $\upset {U}$ is weakly contractible
\end{cor}
\begin{proof}
Each basis element $U$ contains basis elements $V_k$ such that ${\overline V_k}\subseteq V_{k+1}$, and $\bigcup_{k} {\overline V_{k}}=U$.  It follows that
$${\upset U}=\bigcup_{k}{\upset {\overline V_{k}}}=\bigcup_{k}{\upset {V_{k}}}.$$
Every singular sphere $\varphi $ in $\upset U$ has compact image, and so lies in some
${\upset {V_{k}}}$, Thus, by Theorem \ref{2stage}, $\varphi $ represents a trivial element of homotopy in $\upset U$.
\end{proof}

\subsection{Pre-images of compact polyhedral sets} Here we prove Theorem
\ref{inverse}, relating $\Delta ({\upset A})$ to $f^{-1}(\upset A)$
when $A$ is a compact polyhedral subset of $X_1$.

\vskip 5pt
Let $d$ be a length metric on $X_1$. Define $\varphi \colon X_{1}\times (0,1]\to I$ by $\varphi (x,u)=\text{min}\left\{\frac{d(x,A)}{u},1\right\}$. Writing $\varphi _{u}$ for
$\varphi (\cdot,u)$ we note

\begin{enumerate}[(i)]
\item $\varphi _{u}=0$ on $A$;
\item $\varphi _{u}=1$ on $X_{1}-N_{u}(A)$;
\item $0<\varphi _{u}<1$ elsewhere.
\end{enumerate}
\vskip 5pt
Let $B$ be a compact polyhedral subset of $X_2$.
For each $u \in (0, 1]$, define
$\psi _{u} \colon X_{1}\times B\times I\to X_{1}\times B\times I$ by

\[ \psi _{u}(x,y,t)=\begin{cases}
      (x,y,\varphi _{u}(x)) & \text{ if } t\leq \varphi _{u}(x) \\
      (x,y,t) & \text{ if } \varphi _{u}(x)\leq t.
   \end{cases}
\]

\noindent We note that $\psi _{u} $ fixes $A\times B\times I$ and that
$\psi _{u} (x,y,t)=(x,y,1)$ when $x\in X_{1}-N_{u}(A)$ and $y\in B$.
\vskip 5pt
We also note the homotopy between $\psi _{u}$ and the identity map
$h_{u} \colon X_{1}\times B\times I\times [0,1]\to X_{1}\times B\times I$
defined by

\[ h_{u}(x,y,t,v)=\begin{cases}
      (x,y,v\varphi (x)) & \text{ if } t\leq v\varphi _{u}(x) \\
      (x,y,t) & \text{ if } v\varphi _{u}(x)\leq t
   \end{cases}
\]

\noindent and that the image of $\psi _{u}$ is the subset
$\{(x,y,t) \in X_{1}\times B\times I\mid \varphi _{u}(x)\leq t\leq1\}$.
\vskip 5pt
Factoring by the canonical quotient $X_{1}\times B\times I\to X_{1}\ast
B$ we see that, while $\psi _{u}$ does not induce a well-defined map $X_{1}\ast
B\to X_{1}\ast B$, it does induce a map
$\Psi _{u} \colon (X_{1}\ast B)-(X_{1}-A)\to X_{1}\ast B$. Similarly, the homotopy $h_u$ induces a homotopy $H_{u} \colon (X_{1}\ast B)-(X_{1}-A))\times [0,1]\to X_{1}\ast B$.
\vskip 5pt
The image of $\Psi _{u}$ is the compact set
$I_{u} \coloneqq \{(1-t)x+ty\in X_{1}\ast B\mid t\geq \varphi _{u}(x)\}$.
\vskip 5pt
\begin{prop} Given $u$ in $(0,1]$ and a neighborhood $M$ of $N_{u}(A)\ast B$ in $X_{1}\ast X_{2}$, we can deform $(X_{1}\ast B)-(X_{1}-A)$ within itself onto the
compact set $I_{u}\subseteq M$ by a deformation that is continuous in the variable $u$.
\end{prop}
By careful choice of the metric $d$ it can be arranged that for $u=\frac{1}{n}$ the above functions are PL, and in particular that $I_{\frac{1}{n}}$ is polyhedral. Thus we get:
\begin{prop} Given $u=\frac{1}{n}$ and a neighborhood $M$ of $N_{u}(A)\ast B$ in $X_{1}\ast X_{2}$, we can deform $(X_{1}\ast B)-(X_{1}-A)$ within itself onto the
compact polyhedral set $I_{\frac{1}{n}}\subseteq M$.
\end{prop}
\vskip 5pt
Since the order relation $P$ is polyhedral, this proposition can be restricted as follows, where we now take $B$ to be  $A'$ (compact and polyhedral by Proposition \ref{poly}):
\vskip 5pt
\begin{cor}\label{polyneigh} Given $u=\frac{1}{n}$ and a neighborhood $M$ of
$\Delta (\upset A)$ in $\Delta ({\mathcal X})$, $f^{-1}(\upset A)$
can be deformed within itself onto a compact polyhedral set $J_{n}\subseteq M$.
\end{cor}
\begin{proof} The set  $f^{-1}(\upset A)$ consists of $\Delta (\upset
A)$ together with line segments ending in $A'$ whose intersections with
$X_{1}$ have been deleted. Here, we write $J_n$ for the relevant part
of  $I_{\frac{1}{n}}$.  \end{proof}
\vskip 5pt
\begin{lemma}\label{delta} The set $\Delta (\upset A)$ is a compact polyhedral subset of $X_{1}*X_{2}$.
\end{lemma}
\begin{proof}
By Lemmas \ref{upcompact} and \ref{poly}, $A'$ is compact and
polyhedral. Thus $C \coloneqq ((A\times A')\bigcap P)\times I$ is
compact and polyhedral. The projection maps $A\times A'\to A$ and $A\times
A'\to A'$ define PL maps $\pi _{A} \colon ((A\times A')\bigcap P)\times \{0\}\to
A$ and $\pi _{A'} \colon ((A\times A')\bigcap P)\times \{1\}\to A'$. The space
$\Delta (\upset A)$ is clearly decomposable as the union of three compact
polyhedra: $C$ and the mapping cylinders of $\pi _{A}$ and $\pi _{A'}$,
where the mapping cylinders are glued to the $0$- and $1$- ends of $C$
in the obvious way. Since mapping cylinders of PL maps between compact polyhedra
are compact polyhedra, and the union of compact polyhedra glued along compact polyhedral subsets is again a compact polyhedron, this completes the proof.
\end{proof}
\vskip 5pt
\begin{thm}\label{inverse}
When $A$ is a compact polyhedral subset of $X_1$, the inclusion map $\Delta (\upset A)
\hookrightarrow f^{-1}(\upset A)$ is a homotopy equivalence.
\end{thm}
\begin{proof}
$\Delta (\upset A)$ is the intersection of the nested sequence of polyhedra $J_n$, each of which is a strong deformation retract of $f^{-1}(\upset A)$ by Corollary \ref{polyneigh}.
Since the inclusion
maps $J_{n+1}\to J_n$ are homotopy equivalences, shape theory (see Section \ref{appendix} for details) implies that $\Delta (\upset A)$ is shape equivalent to each of the polyhedra $J_n$. Moreover, since $\Delta (\upset A)$ is itself polyhedral by Lemma \ref{delta}, this means that $\Delta (\upset A)$ is actually homotopy equivalent to each of the $J_n$, hence also to $f^{-1}(\upset A)$.
\end{proof}

\vskip 5pt

\subsection{Contractibility of pre-images of basis elements}\label{2.1}

The order complex is the subset $\Delta ({\mathcal X}) \subseteq X_{1}*X_{2}$  which
consists of $X_1$, $X_2$, and the union of all segments joining $x$ to $y$ such that $x<y$.
The Comparison Map $f \colon \Delta ({\mathcal X})\to {\mathcal X}$ is defined by
$f(x)=x$ when $x\in X_{1}$ (i.e. when $t=0$)  and $f((1-t)x+ty)=y$ when $t>0$.

As before, we consider the two kinds of basis elements in the Up topology:
$U\in {\mathcal U}_{2}$ and $\upset U$ where $U \in {\mathcal U}_{1}$. To
apply Theorem \ref{mccord} we show that the pre-image under $f$ of each
of these is weakly contractible.
\vskip 5pt
When $U\in {\mathcal U}_{2}$, the set $f^{-1}(U)$ consists of the contractible set
$U\subseteq X_{2}\subseteq X_{1}*X_{2}$, together with half-open segments ending in $U$,
so $U$  is a strong deformation retract of $f^{-1}(U)$ which is therefore contractible.
\vskip 5pt
Now we consider the case where $U\in {\mathcal U}_{1}$. It is convenient to first deal with  $f^{-1}(\upset {\overline U})$, which Theorem \ref{inverse} tells us is homotopy equivalent to
$\Delta (\upset {\overline U})$.

\begin{notation} We write ${\mathcal C}A$ for the topological cone on the space $A$,
i.e. $A\times I\slash A\times \{0\}$.
\end{notation}

\begin{prop}\label{cone} $\Delta (\upset {\overline U})$ is homotopy equivalent to ${\mathcal C}\overline U'$, and is therefore contractible.
\end{prop}
\begin{proof} Let $g\colon\Delta (\upset {\overline U})\to {\mathcal
C}\overline U'$ be the map which takes $\overline U$ to the cone point $p$, is the
identity on $\overline U'$, and maps the point  $(1-t)x+ty$ to the
segment  $(1-t)p+ty$ in the cone. By Lemma \ref{delta} this is a map between compact polyhedra. Consider $g^{-1}((1-t)p+ty)$. When
$t=0$ this pre-image is the contractible set $\overline U$; when $t=1$
this pre-image is a single point; when $0<t<1$ this pre-image is homeomorphic to $\{x\in
{\overline U}\mid x<y\}$ which has trivial shape (actually, is contractible) by Property A3.
Thus $g$ is a cell-like map, and is therefore a homotopy equivalence by Theorem \ref{lacher2}.
\end{proof}
\vskip 5pt
\begin{cor}\label{contractible} $f^{-1} (\upset {\overline U})$ is contractible.
\end{cor}
\vskip 5pt
\begin{cor}\label{contractible2} For basis elements $U$, $f^{-1} (\upset {U})$ is contractible.
\end{cor}
\begin{proof} The proof is similar to that of Corollary \ref{open}. It shows that $f^{-1} (\upset {U})$ is weakly contractible. Since it is an open subset of a polyhedron, it is actually contractible.
\end{proof}
This completes the proof of Theorem \ref{main} for the case ${\mathcal R}=[2]$.

\section{The special case ${\mathcal R}=[3]$}\label{3case}
\vskip 5pt
\begin{notation} In this section, when $A\subseteq X_{1}$ we write $A'$ rather than $A^{(2)}$ and $A''$ rather than $A^{(3)}$.
\end{notation}
\vskip 5pt
\subsection{Weak contractibility of basis elements}\label{3weak}
Here ${\mathcal X}=X_{1}\coprod X_{2}\coprod X_{3}$ and $U\subseteq X_i$ is a basis element. We are to show that $\uparrow U$ is weakly contractible. If $U\subseteq X_3$ this is trivial, and if $U\subseteq X_2$ this follows from what was done in Section \ref{2case}. So we assume $U\subseteq X_1$.
\vskip 5pt
As before, it is easier to begin with $\overline U$. We consider an arbitrary map
$$g \colon (B,\partial B)\to (\upset {\overline U}, {\overline U}\coprod {\overline U}').$$
\noindent where $B$ abbreviates $B^n$. We write $C:=g^{-1}({\overline U}\coprod {\overline U}')$, a closed subset of $B$ containing $\partial B \subseteq C$. We wish to produce maps $h''$ and $h$ from $B$ to $\upset {\overline U}$ satisfying:
\begin{enumerate}[(i)]

\item $h''=h=g$ on $C$;
\item $h''$ maps $B-C$ into ${\overline U}''$;
\item $h$ maps $B$ into ${\overline U}\coprod {\overline U}'$;
\item on $B-C$, $h''$ covers $h$;
\item $h''$ is homotopic to $g$ rel $C$.
\end{enumerate}
\noindent Proposition \ref{cover} will then imply that $h''$ is homotopic to $h$ rel $C$
and hence that the map $g$ represents the trivial element of $\pi _{n}(\upset {\overline U}, {\overline U}\coprod{\overline U'})$; hence that group is trivial. By Theorem \ref{2stage} it will follow that $\pi _{n}(\upset {\overline U}, {\overline U})$ is trivial.
As before, we can deduce the same when ${\overline U}$ is replaced by $U$.
\vskip 5pt
The argument here is more complicated than in Section \ref{2case} because the obvious replacement for ${\overline U}\times {\overline U}'$ in the definition of $Q$, namely $({\overline U}\coprod {\overline U}')\times {\overline U}''$, is not polyhedral (with the Up topology). For this reason we must construct the map $h''$ in two stages, one for ${\overline U}$, the other for ${\overline U}'$.
\vskip 5pt
Let $Q_{1}:=({\overline U}\times {\overline U}'')\bigcap P$. Let ${\mathcal A}_{i}$ be an open cover of $Q_{1}$
by sets of diameter $<\frac{1}{i}$, and let ${\mathcal B}_i$ be the
refinement given by Theorem \ref{hu2}. Let $4\lambda _i$ be a Lebesgue
number for ${\mathcal B}_i$.
\vskip 5pt
We write $C_{1}:=g^{-1}({\overline U})$; this is a closed subset of
$C$. For each $x\in B-C$ we choose a point $r(x)\in C$ realizing $d(x,C)$. Define $Y:=\{x\in B-C\mid d(x,C)=d(x,C_{1})\}$, a closed subset of $B-C$.
\vskip 5pt
We now proceed as in Section \ref{2.2} replacing $B-C$ by $Y$ and $C$ by $C_1$. The triangulation $K$ and its subcomplex $J$ are replaced by a triangulation\footnote{We assumed, for simplicity, in Section \ref{2.2} that each $I_i$ is a subpolyhedron of $B-C$. Similarly, here we assume that $Y$ is a polyhedral subset of $B-C$. As in the previous case, if it is not polyhedral, one covers it by a slightly larger polyhedral set. Again, we omit the details.} $K_1$ of $Y$ and its subcomplex $J_1$. We define $N_{1}:=J_{1}\bigcup C_{1}$, a neighborhood of $C_1$ in $Y$. We get a map $j_{1}:=(k_{1},k_{1}'')$; with components
$k_{1}:J_{1}\to {\overline U}$ and $k_{1}'':J_{1}\to {\overline U}''$. Both of these maps are extended to agree with $g$ on $C_1$. As before, the extended function $j_{1}:N_{1}\to Q_{1}$ is continuous.
\vskip 5pt
The required maps $h$ and $h''$ are defined to agree with $k_{1}$ and $k_{1}''$ respectively on $N_1$. Then $h''$ covers $h$ and $h=h''=g$ on $C_1$. As in Section \ref{2.2} we may assume
$h''|N$ and $g|N$ are homotopic rel $C_1$. The extension of $h$ and $ h''$ to the rest of $K_1$ maintaining the properties (i)-(v) is achieved as in that section. At this point, the Homotopy Extension Property implies that
$g:B\to \upset {\overline U}$ is homotopic rel $C$ to a map $g_1$, where $g_{1}|Y$ has the desired property of covering a map into ${\overline U}$.
\vskip 5pt
Now comes the part that is different to Section \ref{2.2}. We wish to alter $g_1$ rel
$Y\bigcup C$ to have the desired covering property on the rest of $B-C$. The set
$Z:= B-(C\bigcup Y)$ is an open polyhedral subset of $B$ whose frontier is the compact set ${\overset \bullet Z}:=\text{\rm fr}_{B}(Z)$. We note that $C-C_{1}\subseteq {\overset \bullet Z}$.
\vskip 5pt
\includegraphics[width=10cm]{ag5}
\vskip 5pt
Basically, we repeat what has been described above, with $K_{2}$ (triangulating $Z$),
${\overset \bullet Z}$, $g_1$, and $Q_{2}:=(U'\times U'')\bigcap P$ playing the respective
roles\footnote{The item to note is that ${\overset \bullet Z}$ is playing the role previously played by $C_1$.} of $K_{1}$, $C_1$, $g$ and $Q_{1}$. There is a subcomplex $J_{2}$ of $K_{2}$, and a neighborhood $N_{2}$ of ${\overset \bullet Z}$ in $\overline Z$ playing the role previously played by $J_1$ and $N_1$. We get a map $j_{2}=(k_{2}',k_{2}'')$ with components  $k_{2}':J_{2}\to U'$ and $k_{2}'':J_{2}\to U''$.
The construction (imitating what is given in detail in Section \ref{2.2})  means that this map $j_{2}$ extends continuously to agree with $j_{1}$ on ${\overset \bullet Z}$. The extension to the rest of $Z$ works as before. Thus we get maps  $h''$ and $h$ as required.  This shows that  $g$ is homotopic, rel $\partial B$, to a map whose image misses ${\overline U}''$.
Summarizing:
\vskip 5pt
\begin{thm}For all $n$, $\pi _{n}(\upset {\overline U}, {\overline U}\coprod{\overline U'})$ is trivial.
\end{thm}
\vskip 5pt
As explained above, this implies that $\pi _{n}(\upset {\overline U}, {\overline U})$ is trivial, hence also $\pi_{n}(\upset {\overline U})$. Just as in Section \ref{2.2} this leads to:
\begin{cor}\label{open3}
The set $\upset {U}$ is weakly contractible.
\end{cor}
\vskip 5pt
\begin{rem} It was important to handle ${\overline U}$ before ${\overline
U}'$ since we needed $C_1$ to be compact. This is a simple instance of
what is a more serious issue in the general case, below, where we need
a careful ordering of $\mathcal R$ in order to ensure that at each stage
we are dealing with a compact set.
\end{rem}

\subsection{Contractibility of pre-images of basis elements}\label{3.1}

As a special case of what was defined in Section \ref{order}, recall that the order complex is the set $\Delta ({\mathcal X}) \subseteq X_{1}*X_{2}*X_{3}$
consisting of all points $z=\sum\limits_{i=1}^{3}t_{i}x_{i}$ such that when
$i,j\in \text{supp }z$ then $x_{i}<x_{j}$ in $\mathcal X$. In particular, every $x_i$ lies in this set. The Comparison Map $f \colon \Delta ({\mathcal X})\to {\mathcal X}$ is defined by
$f(z)=\text{max }\{x_{i}\mid i\in \text{supp }(z)\}$.
\vskip 5pt

We first consider $U\subseteq X_1$. We describe the set $f^{-1}(\upset {\overline U})\subseteq X_{1}*X_{2}*X_{3}$ in some detail. Certainly it  includes $\Delta (\upset {\overline U})$. Imitating what was done in Section \ref{2.1}, we let $R^{o}$ be the union of a
certain collection of half-open segments, where the missing point of
each segment is its initial point as measured by the poset ${\mathcal
R}=[3]$. These segments are described as follows. In
every case it is to be understood that $x_1<x_2<x_3$ whenever these
comparisons make sense\footnote{{\it Notation:} $[x,y]$ stands for all the points $(1-t)x+ty$ on the segment joining $x$ to $y$.}:

\begin{enumerate}[(i)]
\item $(x_{1}, x_{2}]$ where $x_{1}\notin {\overline U}, x_{2}\in {\overline U}^{(2)}$, and $x_{2}$ is not comparable to any member of ${\overline U}^{(3)}$;
\item $(x_{1}, x_{3}]$ where $x_{1}\notin {\overline U}, x_{3}\in {\overline U}^{(3)}$, and there is no $x_{2}\in {\overline U}^{(2)}$ with $x_1<x_2<x_3$;
\item $(x_{1}, [x_{2}, x_{3}]]$ where $x_{1}\notin {\overline U}, x_{2}\in {\overline U}^{(2)}$, and $x_{3}\in {\overline U}^{(3)}$;
\item $([x_{1}, x_{2}], x_{3}]$ where $x_{1}\notin {\overline U}$, $x_{2}\notin {\overline U}^{(2)}$, and $x_{3}\in {\overline U}^{(3)}$;
\item $(x_{2}, x_{3}]$ where $x_{2}\notin {\overline U}^{(2)}, x_{3}\in {\overline U}^{(3)}$, and there is no $x_{1}\in {\overline U}$ with $x_1<x_2$.
\end{enumerate}

\vskip 5pt
Then
$$f^{-1}(\upset {\overline U})=\Delta (\upset {\overline U}) \cup R^{o}.$$
\vskip 5pt
The set $X_{2}\coprod X_{3}$ is a sub-poset of $\mathcal X$, and we denote
its order complex by $\Delta _{23}$. Recall that $\Delta _{23}$ consists
of copies of $X_2$ and $X_3$ together with a segment\footnote{In all
such expressions, the sum of coefficients is understood to be $1$, and a
segment such as this will require an obvious adjustment when $t_{2}=0$ or
$1$.} $t_{2}x_{2}+t_{3}x_{3}$ joining $x_2$ to $x_3$ whenever $x_{2}<x_3$.
\vskip 5pt
This gives rise to another poset ${\mathcal Y} \coloneqq X_{1}\coprod
\Delta _{23}$ with the partial ordering $x_{1}< t_{2}x_{2}+t_{3}x_{3}$
when $x_1<x_2<x_3$; $\mathcal Y$ is given the Up topology. We denote
the corresponding Comparison Map by $g \colon \Delta ({\mathcal Y})\to
{\mathcal Y}$. Then, as subsets of $X_{1}\ast X_{2}\ast X_{3}$,
we have

$$f^{-1}(\upset {\overline U})=g^{-1}(\upset {\overline U}).$$

Since $g^{-1}(\upset {\overline U})$ is contractible by Proposition \ref{contractible}, we conclude that
$f^{-1}(\upset {\overline U})$ is contractible.
\vskip 5pt
Next we consider the case where $U\subseteq X_2$. We have the Comparison Map
$f_{23} \colon \Delta _{23}\to X_{2}\coprod X_{3}$. By Proposition \ref{contractible}, $f_{23}^{-1}({\upset {\overline U}})$ is contractible. But $f^{-1}({\upset {\overline U}})$ is larger in general, as it includes deleted segments and $2$-simplexes having a vertex in $f_{23}^{-1}({\upset {\overline U}})$, the deletion being the (non-empty) part lying in $X_1$. However, as before, $f_{23}^{-1}({\upset {\overline U}})$ is a strong deformation retract of $f^{-1}({\upset {\overline U}})$, so the latter is contractible.
\vskip 5pt
Finally, there is the case where $U\subseteq X_3$. In that case, just
as in the previous paragraph, $\upset {\overline U}$ is a strong
deformation retract of $f^{-1}({\upset {\overline U}})$, and therefore
the latter is contractible. As in Corollary \ref{contractible2} we conclude:
\vskip 5pt
\begin{prop}\label{contractible3} For basis elements $U$, the space $f^{-1}(\upset {U})$ is contractible.
\end{prop}

\section{The General Case}\label{general}
\vskip 5pt

We are now dealing with the geometric $M$-poset $\mathcal X$ equipped with a mirror map $\mu :{\mathcal X}\to {\mathcal R}$, where ${\mathcal R}$ is a finite poset.
\vskip 5pt
\subsection{Weak contractibility of basis elements}
In Section \ref{3weak} we showed that when ${\overline U}^{(1)}$ is
the closure of a basis element of $X_1$ then $\uparrow {\overline U}^{(1)}$
is weakly contractible. The method was to consider a map $g \colon
(B,\partial B)\to (\upset {\overline U^{(1)}}, {\overline U^{(1)}}\coprod
{\overline U^{(2)}})$, define $C=g^{-1}({\overline U}^{(1)}\coprod
{\overline U}^{(2)})$ and $C_{1}=g^{-1}({\overline U}^{(1)})$. We picked
a closed subset $Y$ of $B-C$ associated with $C_1$, and first adjusted
$g$ on $Y$. Then we made a further adjustment on $Z=B-(C\bigcup Y)$.
\vskip 5pt
If we were dealing with the case ${\mathcal R}=[4]$ we would do the
same thing with one extra step:  $C_{1}$ would be $g^{-1}({\overline
U}^{(1)})$,  $C_{2}$ would be $g^{-1}({\overline U}^{(1)}\coprod
{\overline U}^{(2)})$ and $C$ would be $g^{-1}({\overline U}^{(1)}\coprod
{\overline U}^{(2)}\coprod {\overline U}^{(3)})$. There would be subspaces $Y$ and $Z$ associated with $C_1$ and $C_2$. The adjustment of
$g$ would be made first on $Y$, then on $Z$ and finally on the rest of
$B-C$. Thus $g$ would be homotopic, rel $\partial B$, to a map whose image
misses ${\overline U}^{(4)}$.
\vskip 5pt
Enough has been said here to indicate that the same is true for
${\mathcal R}=[m]$; namely, if $U^{(1)}$ is a basis element of $X_1$ then
$\uparrow U^{(1)}$ is weakly contractible.
\vskip 5pt
To get the same result for an arbitrary finite poset $\mathcal R$, we need
to order its elements with care. Let
$${\mathcal R}_{0}:=\{r\in {\mathcal R}\mid \text{there is no }s\in {\mathcal R} \text{ with }s<r\}.$$
For $i>0$ let
$${\mathcal R}_{i}:=\{r\in {\mathcal R}\mid \text{the longest chain connecting }
 r \text{ to an element of }{\mathcal R}_{0} \text{ has length }i\}.$$
We choose a total ordering of  $\mathcal R$ so that, for all $i$, every element of ${\mathcal
R}_i$ comes before any element of ${\mathcal R}_{i+1}$. We call this the
{\it useful ordering} to distinguish it from the given partial ordering
of $\mathcal R$. The useful ordering ensures a key point: that for any $r$,
$\coprod \{X_{s}\mid s \text{ precedes }r\text{ in the useful ordering}\}$ is closed in the Up topology on ${\mathcal X}$.
\vskip 5pt
We are to show that, for any basis element $U^{(s)}$ in $X_s$, $\upset
{\overline U}^{(s)}$ is weakly contractible. There is no loss of generality in
assuming $s=m$ where $m$ is the minimal element in the useful ordering. Let $M$ be the maximal element in the useful ordering. We consider a map
$g \colon (B,\partial B)\to (\upset {\overline U}^{(m)}, \upset {\overline U}^{(m)}-{\overline U}^{(M)})$. If the elements of the useful ordering are taken in ascending order, then an obvious generalization of the above
discussion for ${\mathcal R}=[3]$ and ${\mathcal R}=[4]$ adjusts $g$, rel
$\partial B$, to a map whose image misses ${\overline U}^{(M)}$. Proceeding
by induction we get:
\begin{thm}For all $n$, $\pi _{n}(\upset {\overline U}^{(m)}, {\overline U}^{(m)})=0$.
\end{thm}
\vskip 5pt
Just as in Section \ref{2.2} this leads to:
\begin{cor}\label{opengeneral} The set $\upset {U^{(m)}}$ is weakly contractible.
\end{cor}

\subsection{Contractibility of pre-images of basis elements}\label{4.1}
The Comparison Map $f:\Delta ({\mathcal X})\to {\mathcal X}$ is, as before,
$f(z)=\text{max}\{x_{i}\mid i\in \text{ supp}(z)\}$. Let $U\subseteq X_r$ where $r\in {\mathcal R}$.
\vskip 5pt
\begin{prop}\label{preimagegeneral} $f^{-1}(\uparrow {\overline U})$ is contractible.
\end{prop}
\begin{proof}
We imitate Section \ref{3case}, inducting on the cardinality of $\mathcal R$. When
$\mathcal R$ is a singleton the proposition is trivially true. When $\mathcal R$ has two elements the proposition has been proved. We assume $\mathcal R$ has at least three elements.
\vskip 5pt
There are two cases
\vskip 5pt
First assume $r$ is a minimal element of $\mathcal R$. Then
$$f^{-1}(\uparrow {\overline U})=\Delta (\uparrow {\overline U}) \bigcup
R^{o}$$ where $R^o$ is the union of deleted simplexes, each having some
of its vertices outside $\uparrow {\overline U}$ and the remaining vertices
(at least one) in $\uparrow {\overline U}$, the deletions being the
faces lying outside $\uparrow {\overline U}$.  Now, ${\mathcal Z}:={\mathcal
X}-X_r$ is a sub-poset. From it we can construct a new poset ${\mathcal
Y}:=X_{r}\coprod \Delta ({\mathcal Z})$ where the details imitate the
construction given explicitly in Section \ref{3.1}. The Comparison Map
is $g:\Delta ({\mathcal Y})\to {\mathcal Y}$. Then $$f^{-1}(\uparrow
{\overline U})=g^{-1}(\uparrow {\overline U}).$$ By induction and
Corollary \ref{contractible}, $g^{-1}(\uparrow {\overline U})$ is
contractible, so we conclude that $f^{-1}(\uparrow {\overline U})$
is contractible.

\vskip 5pt
In the other case, $r$ is not minimal. Then there is a proper
sub-$M$-poset $\uparrow X_r$ indexed by a proper subset of $\mathcal
R$. By induction on the cardinality of $\mathcal R$, $g^{-1}(\uparrow {\overline
U})$ is contractible, where $g$ denotes the Comparison Map for  $\uparrow
X_r$. Then $$f^{-1}(\uparrow {\overline U})=g^{-1}(\uparrow {\overline
U})\bigcup R^{o}$$ where $R^o$ is the union of deleted simplexes, each
having some of its vertices outside $\uparrow {\overline U}$ and at least one vertex in $\uparrow{\overline U}$, the deletions being the faces lying  outside $\uparrow {\overline U}$. These deleted simplexes deform to $g^{-1}(\uparrow {\overline U})$. So  $f^{-1}(\uparrow {\overline U})$ is contractible.
\end{proof}
\vskip 5pt
As in Corollary \ref{contractible2} we conclude:
\begin{cor}
For basis elements $U$, $f^{-1}(\uparrow U)$ is contractible
\end{cor}
\vskip 5pt
This completes the proof of Theorem \ref{main}.

\section{Appendix on the shape theory used in this paper.}\label{appendix}

Shape theory is a variant of homotopy theory which works better than
regular homotopy theory when the spaces are not locally nice. Here we
give a brief review with the sole purpose of  amplifying our use of the
theory in Section \ref{2.1}.

\vskip 5pt
Shape theory considers a compact metrizable space $X$ by dealing instead with an arbitrary  inverse sequence of compact polyhedra and maps whose inverse limit is homeomorphic to $X$.
\vskip 5pt

$$
\xymatrix{
&X_{1}
&X_{2}\ar[l]^{h_{1}}
&X_{3}\ar[l]^{h_{2}}
&\ar[l]^{h_{3}}{\cdots}\\
}
$$

\vskip 5pt
In the following infinite diagram of space and maps, the spaces
are compact polyhedra. We denote the inverse limit of the top line
[resp. bottom line] by $X$ [resp.$Y$]. The outer squares are assumed to
be homotopy commutative. If there exist diagonal maps making the entire
diagram homotopy commutative then $X$ and $Y$ are said to be {\it shape
equivalent}\footnote{This is not the general definition but is enough
for our purposes. For more details on what is discussed in this Appendix, see \cite{Gu}.}
\vskip 5pt
$$
\xymatrix{
&X_{1}\ar[d]
&X_{2}\ar[l]\ar[d]
&X_{3}\ar[l]\ar[d]
&\ar[l]{\cdots}\\
&Y_{1}
&Y_{2}\ar[l]\ar[ul]
&Y_{3}\ar[l]\ar[ul]
&\ar[l]\ar[ul]\cdots\\
}
$$

\vskip 5pt

Now consider the special case in which all the maps $h_i$ are homotopy equivalences. Then we have the following homotopy commutative diagram, where $h_{i}^{-1}$ stands for a homotopy inverse of $h_i$:
$$
\xymatrix{
&X_{1}\ar[d]_{id}
&X_{1}\ar[l]_{id}\ar[d]^{h_{1}^{-1}}
&X_{1}\ar[l]_{id}\ar[d]^{(h_{1}h_{2})^{-1}}
&\ar[l]_{id}{\cdots}\\
&X_{1}
&X_{2}\ar[l]^{h_{1}}\ar[ul]_{h_{1}}
&X_{3}\ar[l]^{h_{2}}\ar[ul]_{h_{1}h_{2}}&\ar[l]^{h_{3}}{\cdots}\\
}
$$

We conclude that in this case $X$ is shape equivalent to the compact
polyhedron $X_1$. Moreover, in the application in Section \ref{2.1}, $X$
itself known to be a compact polyhedron. It is a basic theorem of shape
theory that {\it two polyhedra are shape equivalent if and only if they
are homotopy equivalent.}

\vskip 5pt
In the context of Proposition \ref{inverse} this discussion explains why $\Delta(\overline U)$ is homotopy equivalent to each $P_n$ and hence to $f^{-1}(\upset {\overline U})$.

\section{Appendix on the discrete case and McCord's work.}\label{discrete}
\vskip 5pt

A quick review of the present paper shows that the proof of the Comparison
Theorem \ref{main} becomes very easy in the case when each $X_r$ is
discrete. Confining ourselves to the special case dealt with in Section
\ref{2case}, the relevant version of Proposition \ref{inverse} becomes almost
trivial, since the issue of limit points (which required a shape theoretic
argument) is not present, and the need for Proposition \ref{cone} is
gone since pre-images of basic sets $\upset x$ are cones. The issues
dealt with in Section \ref{2.2} become trivial because Proposition \ref{cover}
implies that each singleton $\{x\}$ is a strong deformation retract of
the set $\upset x$.

\vskip 5pt

McCord's Theorem $3$ in \cite{Mc2} implies our Comparison Theorem \ref{main}
in this discrete case, and his proof is indicated by the contents of the
previous paragraph. It should be added, however, that his definition of
$\Delta ({\mathcal X})$ is superficially different from ours. In his work,
the role of $\Delta ({\mathcal X})$ is played by $|K({\mathcal X})|$,
where $K({\mathcal X})$ is the classical order complex defined by the poset
$\mathcal X$, namely: simplexes are finite chains in the poset. However,
it is not hard to see that  $|K({\mathcal X})|$ and $\Delta ({\mathcal X})$
are the same in our setting when $\mathcal X$ is discrete.
\vskip 5pt

McCord uses the ``Down topology" rather than the ``Up topology" but
that is of no consequence as the two are dual to one another
and order complexes are preserved under duality.

\vskip 5pt
It should be added that McCord's Theorem $3$ covers cases outside the setting of the present paper.
\section{Appendix on homotopy equivalence}
\vskip 5pt
In Theorem \ref{main} we showed that for geometric $M$-posets $\mathcal X$ the Comparison Map $f \colon \Delta ({\mathcal X})\to {\mathcal X}$ is a weak homotopy equivalence. The Whitehead Theorem implies that $f$ is a homotopy equivalence if and only if $\mathcal X$ has the homotopy type of a $CW$ complex. In this appendix we show that topological posets $\mathcal X$ in an important class do not have the homotopy type of a $CW$ complex
\vskip 5pt
Let $\mathcal X$ be a topological poset, not necessarily geometric. Define
an equivalence relation on $\mathcal X$ by $p\sim q$ if and only if there
is a finite sequence $(x_i)$ in $\mathcal X$ with $x_0=p$, $x_n=q$,
and, for each $i\geq 0$, either $x_i \leq x_{i+1}$ or $x_i \geq x_{i+1}$.
Clearly, each equivalence class lies in a path component. We say that
$\mathcal X$ has {\it discrete type} if each equivalence class is an
entire path component\footnote{The topological poset $\mathcal X$ has
an underlying discrete poset which we denote by ${\mathcal X}_{\delta}$,
equipped with the Up topology. Comparing  ${\mathcal X}_{\delta}$ with its
(classical) order complex, one easily sees that each equivalence class
is a path component of ${\mathcal X}_{\delta}$. This explains the term
``discrete type".}
\vskip 5pt
\begin{example} For each $n$ the real Grassmann Poset ${\mathcal G}_n$ has discrete type.
\end{example}
\begin{lemma}\label{all} Each equivalence class is a  closed subset of $\mathcal X$.
\end{lemma}
\begin{proof} The relation $P$ in the definition of topological poset is closed in ${\mathcal X}\times \mathcal X$, so the condition of not being related is an open condition. It follows that
every limit point of an equivalence class lies in that equivalence class.
\end{proof}
\vskip 5pt
For any space $Y$ the space of path components, $\pi _{0}(Y)$, occurs as a quotient space in a natural way. When $Y$ is a $CW$ complex, this space is discrete.
\vskip 5pt
\begin{thm}\label{ok} Let $\mathcal X$ have discrete type. Then $\mathcal X$ has the homotopy type of a $CW$ complex if and only if each path component is contractible, and the space of path components $\pi _{0}({\mathcal X})$ is discrete.
\end{thm}
\begin{proof} Let $h: K\to {\mathcal X}$ be a homotopy equivalence, where $K$ is a $CW$ complex. Suppose there is a homotopy inverse $g$ for $h$. The fact
that  $g$ is continuous shows that if $x<y$ then $g(y)=g(x)$; hence $g$ is
constant on equivalence classes (i.e. on path components). The map $g$ induces
a continuous bijection $\pi _{0}({\mathcal X})\to \pi _{0}(K)$, so $\pi _{0}({\mathcal X})$ is discrete. And since $h\circ g$ is homotopic to the identity map, each path component of $\mathcal X$ must be contractible. Conversely, the hypotheses imply that $\mathcal X$ has the homtopy type of a discrete space.
\end{proof}
\begin{rem} It follows from Theorem \ref{ok} that equivalence classes
are closed. We include Lemma \ref{all} because this is true whether or
not $\mathcal X$ has discrete type. We wonder whether Theorem \ref{ok}
holds for all topological posets.
\end{rem}
\vskip 5pt
\begin{ack} We thank Laura Anderson for telling us about this problem
and for educating us on its connection with tropical phased matroids.
\end{ack}

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