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

Based on a theorem of Lov\'{a}sz \cite{Lovasz_1967} and an own observation \cite[Theorem 5]{aCampo_2018}, the author \cite{aCampo_toappear_1,aCampo_toappear_2} has worked about homomorphism sets under the aspect:{\em  What is it in the structure of finite poests $R$ and $S$ that results in $\# {\cal H}(P,R) \leq \# {\cal H}(P,S)$ for every finite poset $P$?} (${\cal H}(A,B)$ denotes the set of homomorphisms from the poset $A$ to the poset $B$.) Three nontrivial examples for such pairs of posets are shown in Figure \ref{figure_1}. More can be found in \cite{aCampo_toappear_1}, where a detailed survey over the research about homomorphism sets and its results is also presented.

\begin{figure}
\begin{center}
\includegraphics[trim = 70 710 200 70, clip]{Cancel_Abb_1.pdf}
\caption{\label{figure_1} Three examples for posets $R$ and $S$ with $\# {\cal H}(P,R) \leq \# {\cal H}(P,S)$ for every finite poset $P$ \cite{aCampo_toappear_1}.}
\end{center}
\end{figure}

The systematic examination of the phenomenon ``$\# {\cal H}(P,R) \leq \# {\cal H}(P,S)$ for every finite poset $P$'' in \cite{aCampo_toappear_1,aCampo_toappear_2} is based on the concept of the {\em strong Hom-scheme from $R$ to $S$}. In the case of existence, such a Hom-scheme defines a one-to-one mapping $\rho_P : {\cal H}(P,R) \rightarrow {\cal H}(P,S)$ for every finite poset $P \in  \myfr{P} _r$, where $ \myfr{P} _r$ is a representation system of the non-isomorphic finite posets. Postulating regularity conditions for the way, how $\rho_P$ maps homomorphimss from ${\cal H}(P,R)$ to ${\cal H}(P,S)$, gives rise to the definition of the {\em strong G-scheme} and the {\em strong I-scheme}. The relations $R \sqsubseteq S$ (there exists a strong Hom-scheme from $R$ to $S$), $R \sqsubseteq_G S$ (existence of a strong G-scheme), and $R \sqsubseteq_I S$ (existence of a strong I-scheme), turn out to be partial order relations on $ \myfr{P} _r$.

The present paper deals with calculation and cancellation rules for strong Hom-, G-, and I-schemes. Let ${\preceq} \in \{ \sqsubseteq, \sqsubseteq_G, \sqsubseteq_I  \}$. Our questions are:
\begin{itemize}
\item For finite posets $R, S$ with $R \preceq S$, what is the relation between the duals of $R$ and $S$, and how is $R \preceq S$ connected with ${\cal H}(Q,R) \preceq {\cal H}(Q,S)$ for every $Q \in  \myfr{P} $?
\item Let $R_1, R_2, S_1, S_2$ be finite posets with $R_1 \preceq S_1$ and $R_2 \preceq S_2$. Does this imply $R_1 \odot R_2 \preceq S_1 \odot S_2$, where $ \odot $ denotes the poset-operator direct sum, ordinal sum, or product?
\item Let $Q, R, S$ be finite posets with $ Q \odot R \preceq Q \odot S$, $\odot$ as above. When does $R \preceq S$ hold?
\end{itemize}

After the preparatory Sections \ref{sec_notation} and \ref{sec_connectivity}, the required apparatus about Hom-schemes is recalled from \cite{aCampo_toappear_1,aCampo_toappear_2} in Section \ref{sec_EV_HomS}. Additionally, it is shown in Proposition \ref{prop_zshP_reicht} that a (strong) Hom-scheme, G-scheme, or I-scheme only defined for connected finite posets can be extended to all finite posets.

Section \ref{sec_calcrules} containes the calculation rules. They work without exceptions for strong Hom-schemes and strong G-schemes, but for strong I-schemes, we did not prove a calculation rule for ordinal sums and products.

The Sections \ref{sec_cancel_dirsum}, \ref{sec_cancel_ordsum}, and \ref{sec_cancel_product} contain the cancellation rules. For the direct sum of posets (Section \ref{sec_cancel_dirsum}), the cancellation rules work with no ifs or buts, but for ordinal sums (Section \ref{sec_cancel_ordsum}) and products (Section \ref{sec_cancel_product}), we need additional assumptions for strong I-schemes and partly also for strong G-schemes. Most of the proofs are quite technical, but the ansatz is similar in many of them, as explained in Section \ref{sec_ansatz_cancel}. However, there remains a gap: We did not succeed in proving a cancellation rule for $Q \oplus R \sqsubseteq Q \oplus S$.

\section{Basics and Notation} \label{sec_notation}

For a given set $X$, a subset of $X \times X$ is called a {\em (binary) relation} $R$ on $X$ and $X$ is called the {\em carrier} of $R$. The {\em induced relation} on $A \subseteq X$ is $R \cap ( A \mytimes A )$. We write $x R y$ for $(x,y) \in R$. The {\em dual relation} $R^d$ is defined by $x R^d y \Leftrightarrow y R x$ for all $x, y \in X$. A reflexive, antisymmetrical, and transitive relation $R$ on $X$ is called a {\em partial order relation}, and the pair $P = (X,R)$ is called a {\em partially ordered set} or simply a {\em poset}. For an reflexive and antisymmetrical relation (hence, in particular, for a partial order relation), we use the symbol $\leq$. ``$x < y$'' means ``$x \leq y$ and $x \not= y$''.

A subset $A \subseteq X$ is called a {\em chain} iff $x R y$ or $y R x$ for all $x, y \in A$; the cardinality of a chain is called its {\em length}. If a relation $R$ contains a finite longest chain, the length $h_R$ of such a chain is called the {\em height} of $R$.

For relations $R$ and $S$ with disjoint carriers $X$ and $Y$, their {\em direct sum} $R + S$ is defined as $R \cup S$, and their {\em ordinal sum} as $R \oplus S \equiv R \cup S \cup (X \times Y)$. Given posets $P = (X, \leq_P ) $ and $Q = (Y, \leq_Q)$, their product $P \mytimes Q \equiv (X \mytimes Y, \leq_{P \mytimes Q}) $ on $X \mytimes Y$ is defined by $(x_1,y_1) \leq_{P \mytimes Q} (x_2, y_2) $ iff $ x_1 \leq_P x_2$ and $y_1 \leq_Q y_2$.

For relations $R$ and $S$ on $X$ and $Y$, respectively, a mapping $\xi : X \rightarrow Y$ is called a {\em homomorphism}, iff $\xi(x) S \xi(y)$ holds for all $x, y \in X$ with $x R y$. A homomorphism $\xi$ is called {\em strict} iff it additionally fulfills $( x R y$ and $x \not= y ) \Rightarrow ( \xi(x) S \xi(y)$ and $\xi(x) \not= \xi(y) )$ for all $x, y \in X$, and it is called an {\em embedding} iff $\xi(x) S \xi(y) \Rightarrow x R y$ for all $x, y \in X$. Finally, an embedding is called an {\em isomorphism} iff it is onto. For posets $P$ and $Q$, we let ${\cal H}(P,Q)$ and ${\cal S}(P,Q)$ denote the set of homomorphims from $P$ to $Q$ and the set of strict homomorphims from $P$ to $Q$, respectively. $P \simeq Q$ indicates isomorphism. We equip homomorphism sets with the ordinary pointwise partial order.

$ \myfr{P} $ is the class of all {\em non-empty} finite posets, and $ \myfr{P} _r$ is a representation system of the non-isomorphic posets in $ \myfr{P} $. (Without loss of mathematical substance, we avoid repeated trivial distinctions of cases by excluding the empty poset.)

For a poset $P$ with carrier $X$, we use the notation $x \in P$ instead of $x \in X$, and for posets $P$ and $Q$ with carriers $X$ and $Y$, respectively, we write $\xi : P \rightarrow Q$ instead of $\xi : X \rightarrow Y$ for a homomorphism $\xi \in {\cal H}(P,Q)$.

Given a relation $R$ on $X$, we define for $A \subseteq X$
\begin{align*}
{\downarrow \,} A & \equiv \mysetdescr{ y \in X }{ \exists \; a \in A \; \mytext{:} \; y R a }, \\
{\downarrow^{^{\!\!\!\circ}}} A & \equiv ( {\downarrow \,} A ) \setminus A, \\
{\uparrow \,} A & \equiv \mysetdescr{ y \in X }{ \exists \; a \in A \; \mytext{:} \;  a R y }, \\
{\uparrow_{_{_{\!\!\!\circ}}}} A & \equiv ( {\uparrow \,} A ) \setminus A.
\end{align*}
For $x \in X$, we write ${\downarrow \,} x$ and ${\downarrow^{^{\!\!\!\circ}}} x$ instead of ${\downarrow \,} \{x\}$ and ${\downarrow^{^{\!\!\!\circ}}} \{x\}$, respectively, and correspondingly ${\uparrow \,} x$ and ${\uparrow_{_{_{\!\!\!\circ}}}} x$. If required, we label the arrows with the relation they are referring to. $A \subseteq X$ is called an {\em upset} iff $A = {\uparrow \,} A$.

Additionally, we use the following notation from set theory:
\begin{align*}
\myNk{0} & \equiv  \emptyset \\
\myNk{n} & \equiv  \{ 1, \ldots, n \} \mytext{for every} n \in \mathbb{N}.
\end{align*}

We denote the $i$\textsuperscript{th} component of $x \in X_1 \times \cdots \times X_I$ as usual by $x_i$. Due to ${\cal H}( P, Q \times R ) \simeq {\cal H}( P, Q ) \times {\cal H}(P, R)$ for all posets $P, Q, R$, we frequently write $( \xi_1, \xi_2)$ for a homomorphism $\xi \in {\cal H}( P, Q \times R )$. In order to avoid confusion in Section \ref{sec_cancel_product} (where we are dealing with triplets consisting of pairs and sets of pairs), we use additionally the canonical projections $\pi_1$ and $\pi_2$: for every $( a, b ) \in A \times B$, $\pi_1(a,b) \equiv a$ and $\pi_2(a,b) \equiv b$.

For sets $X$ and $Y$, ${\cal A}(X,Y)$ is the set of mappings from $X$ to $Y$. $id_X \in {\cal A}(X,X)$ is the {\em identity mapping}. For a mapping $f \in {\cal A}(X,Y)$, our symbols for the {\em pre-image} of $B \subseteq Y$ and of $y \in Y$ are
\begin{align*}
\myurbild{f}(B) & \equiv \mysetdescr{ x \in X }{ f(x) \in B}, \\
\myurbild{f}(y) & \equiv \myurbild{f} (\{ y\} ).
\end{align*}
For $A \subseteq X$ and $B \subseteq Y$ with $f(X) \subseteq B$, let $f \vert_A : A \rightarrow Y$ and $f \vert^B : X \rightarrow B$ denote the {\em pre-restriction} and {\em post-restriction} of $f$, respectively.

Let $A, B, C, D$ be sets with $A \cap C = \emptyset$. For mappings $f : A \rightarrow B$, $g : C \rightarrow D$, the mapping $ f \cup g : A \cup C \rightarrow B \cup D$ is defined by
\begin{align*}
( f \cup g )(x) & \equiv
\begin{cases}
f(x), & \mytext{if} x \in A; \\
g(x), & \mytext{if} x \in B.
\end{cases}
\end{align*}
Let $A, B$ be sets with $A \subseteq B$. For $f \in {\cal A}(A,B)$, we define recursively for every $a \in A$
\begin{align*}
f^0(a) & \equiv a, \\
\forall \; i \in \mathbb{N} \; : \; f^{i+1}(a) & \equiv f( f^i( a ) ), \; \mytext{if} \; f^i( a ) \in A.
\end{align*}

Let ${\cal I}$ be a non-empty set, and let $N_i$ be a non-empty set for every $i \in {\cal I}$. The {\em Cartesian product} of the sets $N_i, i \in {\cal I}$, is defined as
\begin{eqnarray*}
\prod_{i \in {\cal I}} N_i & \; \equiv \; &
\mysetdescr{ f \in {\cal A} \big( {\cal I}, \bigcup_{i \in {\cal I}} N_i \big)}{ f(i) \in N_i \mytext{for all} i \in {\cal I} }.
\end{eqnarray*}

\section{Connectivity} \label{sec_connectivity}

\begin{definition} \label{def_connected}
Let $P \in  \myfr{P} $, $A \subseteq P$, and $x, y \in A $. We say that $x$ and $y$ are {\em connected in $A$}, iff there are $z_0, z_1, \ldots , z_L \in A$, $L \in \mathbb{N}_0$, with $x = z_0$, $y = z_L$ and $ z_{\ell-1} < z_\ell$ or $z_{\ell-1} > z_\ell$ for all $\ell \in \myNk{L}$. We call $z_0, \ldots , z_L$ a {\em zigzag line connecting $x$ and $y$}. We define for all $A \subseteq P$, $x \in A $, $B \subseteq A$
\begin{align*}
\gamma_A(x) & \equiv \mysetdescr{ y \in A }{ x \mytext{and} y \mytext{are connected in} A}, \\
\gamma_A(B) & \equiv \bigcup_{ b \in B } \gamma_A(b).
\end{align*}
\end{definition}
The sets $\gamma_P(x), x \in P$, are called the {\em connectivity components} of $P$. Every poset  is the direct sum of its connectivity components. A poset $P \in  \myfr{P} $ is {\em connected} iff $\gamma_P(x) = P$ for an/all $x \in P$, and a non-empty subset $A \subseteq P$ is called {\em connected (in $P$)} iff $\gamma_A(x) = A$ for an/all $x \in A$. For posets $P$ and $Q$, $P$ connected, the image $\xi(P)$ of $P$ under a homomorphism $\xi \in {\cal H}(P,Q)$ is connected in $Q$; in particular, $\xi(P)$ is a subset of a single connectivity component of $Q$. In consequence, if $P$ is a connected poset, then for all disjoint posets $R$ and $S$
\begin{align*}
{\cal H}(P, R + S) & \simeq {\cal H}(P,R) + {\cal H}(P,S),
\end{align*}
and correspondingly also for ${\cal S}(P, R + S )$. Furthermore, for posets $P$ and $Q$, every homomorphism $\xi \in {\cal H}(P,Q)$ can be decomposed into its restrictions to the connectivity components $K_i$ of $P$, $i \in {\cal I}$:
\begin{align*}
\xi & = \bigcup_{i \in {\cal I}} \xi \vert_{K_i}.
\end{align*}

The following results have been proven by the author \cite{aCampo_2018,aCampo_toappear_1}:
\begin{lemma} \label{lemma_ueber_gamma}
Let $A \subseteq P$. The relation ``connected in $A$'' is an equivalence relation on $A$ with partition $\mysetdescr{\gamma_A(a)}{a \in A}$. For $B \subseteq A \subseteq A' \subseteq P$ we have
\begin{align}
\gamma_A(B) & \subseteq \gamma_{A'}(B), \label{gamma_ABAB} \\
\gamma_A(B) & = \gamma_{\gamma_A(B)}(B). \label{gamma_ggc}
\end{align}
\end{lemma}

The next definition is fundamental also for the recent article:

\begin{definition}
Let $P = (X, \leq)$ be a finite poset, let $Y$ be a set, and let $\xi \in {\cal A}(X,Y)$ be a mapping. We define for all $x \in P$
\begin{align*}
\mygxix{\xi}{x} & \equiv \gamma_{\myurbild{\xi}( \xi(x) )}( x ).
\end{align*}
\end{definition}
The set $\mygxix{\xi}{x}$ is always a connected subset of $P$ with $x \in \mygxix{\xi}{x}$. We have
\begin{align} \label{gxix_gammagxix}
\mygxix{\xi}{x} & = \gamma_{\mygxix{\xi}{x}}(x),
\end{align}
and $\xi$ is strict iff $\mygxix{\xi}{x} = \{x\}$ for all $x \in P$. For a one-to-one mapping $\sigma : Y \rightarrow Z$, we have $\mygxix{\sigma \circ \xi}{x} = \mygxix{\xi}{x}$ for every $\xi \in {\cal A}(X,Y), x \in P$.

We need two additional lemmata in this paper:

\begin{lemma} \label{lemma_G_prod}
Let $P \in  \myfr{P} $, and let $Y$ and $Z$ be non-empty finite sets. For $\xi \in {\cal A}(P, Y \mytimes Z)$ let $\xi_1 \in {\cal A}(P,Y)$ and $\xi_2 \in {\cal A}(P,Z)$ be defined by $( \xi_1(x), \xi_2(x) ) = \xi(x)$ for all $x \in X$. Then, for all $x \in X$,
\begin{align} \label{gxix_produkt}
\mygxix{\xi}{x} & = \gamma_{ \mygxix{\xi_1}{x} \cap \mygxix{\xi_2}{x} }(x).
\end{align}
\end{lemma}

\begin{proof} For every $ (a,b) \in Y \mytimes Z $, we have $\myurbild{\xi}(a,b) = \myurbild{\xi_1}(a) \cap \myurbild{\xi_2}(b)$, which yields $\myurbild{\xi}( \xi(x) ) \subseteq \myurbild{\xi_j}(\xi_j(x))$ for every $x \in P$ and $j \in \myNk{2}$. With \eqref{gamma_ABAB}, we conclude $\mygxix{\xi}{x} \subseteq \mygxix{\xi_1}{x} \cap \mygxix{\xi_2}{x}$, and we get
\begin{equation*}
\mygxix{\xi}{x} \; \stackrel{\eqref{gamma_ggc}}{=} \; \gamma_{\mygxix{\xi}{x}}(x) \;  \stackrel{\eqref{gamma_ABAB}}{\subseteq} \; \gamma_{ \mygxix{\xi_1}{x} \cap \mygxix{\xi_2}{x} }(x).
\end{equation*}
Now let $y \in \gamma_{ \mygxix{\xi_1}{x} \cap \mygxix{\xi_2}{x} }(x)$, and let $z_0, \ldots , z_L$ be a zigzag line connecting $x$ and $y$ in $\mygxix{\xi_1}{x} \cap \mygxix{\xi_2}{x}$. Then $\xi_1(z_\ell) = \xi_1(x)$ and $\xi_2(z_\ell) = \xi_2(x)$ for every $\ell \in \myNk{L} \cup \{ 0 \}$. The zigzag line $z_0, \ldots , z_L$ connects thus $x$ and $y$ in $\myurbild{\xi_1}(\xi_1(x)) \cap \myurbild{\xi_2}(\xi_2(x)) = \myurbild{\xi}(\xi(x))$, and we conclude $y \in \mygxix{\xi}{x}$.

\end{proof}
\begin{lemma} \label{gxix_neu}
For every $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$
\begin{align} \label{gxix_postrestr_oben}
\forall \; B \subseteq R \; : \; \xi(P) \subseteq B \quad & \Rightarrow \quad \mygxix{ \xi \vert^B }{x} = \mygxix{\xi}{x}, \\
\label{gxix_postrestr_unten}
\forall \; A \subseteq P \; : \; \mygxix{\xi}{x} \subseteq A \quad & \Rightarrow \quad \mygxix{ \xi \vert_A }{x} = \mygxix{\xi}{x},
\end{align}
in the partial order induced on $A$.
\end{lemma}
\begin{proof} Due to $\xi(P) \subseteq B$, we have $\myurbild{\left( {\xi \vert^B} \right)}( \xi \vert^B (x)) = \myurbild{\xi}( \xi (x))$, and \eqref{gxix_postrestr_oben} follows. In the case $\mygxix{\xi}{x} \subseteq A$ we have $x \in A$, and $ \xi \vert_A (x)$ is well-defined. Now
\begin{align*}
\mygxix{\xi}{x} & \subseteq A \cap \myurbild{\xi}( \xi(x) ) =
\myurbild{\left( \xi \vert_A \right) }( \xi(x) ) =
\myurbild{\left( \xi \vert_A \right) }( \xi \vert_A(x) ),
\end{align*}
thus $\mygxix{\xi}{x} \stackrel{\eqref{gxix_gammagxix}}{=} \gamma_{\mygxix{\xi}{x}}(x) \stackrel{\eqref{gamma_ABAB}}{\subseteq} \gamma_{\myurbild{\left( \xi \vert_A \right) }( \xi \vert_A(x) )}(x) = \mygxix{\xi \vert_A}{x}$. Furthermore,
\begin{align*}
\mygxix{\xi \vert_A}{x} & =
\gamma_{ \myurbild{ \left( \xi \vert_A \right) }( \xi \vert_A ( x ) )}(x) =
\gamma_{ \myurbild{ \left( \xi \vert_A \right)  }( \xi( x ) )}(x)
\stackrel{\eqref{gamma_ABAB}}{\subseteq}
\gamma_{ \myurbild{ \xi }( \xi( x ))}(x) = \mygxix{\xi}{x}.
\end{align*}

\end{proof}

\section{Hom-Schemes} \label{sec_EV_HomS}

\begin{figure}
\begin{center}
\includegraphics[trim = 75 640 185 65, clip]{Cancel_Abb_2.pdf}
\caption{\label{figure_2} Examples for EV-systems of posets \cite{aCampo_toappear_1}.}
\end{center}
\end{figure}

In in this section, the required definitions and results about Hom-schemes contained in \cite{aCampo_toappear_1,aCampo_toappear_2} are summarized without proofs. In Proposition \ref{prop_zshP_reicht}, a new result is presented.

\begin{definition} \label{def_EVsys}
Let $P$ be a poset. The {\em EV-system} ${\cal E}(P)$ of $P$ is defined as
\begin{align*}
{\cal E}(P) & \equiv \mysetdescr{ ( x, D, U ) }{ x \in P, D \subseteq  {\downarrow^{^{\!\!\!\circ}}} x, U \subseteq {\uparrow_{_{_{\!\!\!\circ}}}} x }.
\end{align*}
We equip ${\cal E}(P)$ with a reflexive and antisymmetrical relation: For all $\mf{a}, \mf{b} \in {\cal E}(P)$ we define
\begin{equation*}
\mf{a} <_+ \mf{b} \quad \equiv \quad \mf{a}_1 \in \mf{b}_2 \; \mytext{and} \; \mf{b}_1 \in \mf{a}_3,
\end{equation*}
and $\leq_+ \; \equiv \; <_+ \cup \; \mysetdescr{(\mf{a},\mf{a})}{\mf{a} \in {\cal E}(P)}$.
\end{definition}

Four examples of EV-systems are shown in Figure \ref{figure_2}. ``EV'' reminds of the exploded view drawings in mechanical engineering: just as the EV-system ${\cal E}(P)$ does with the points of a poset $P$, the exploded view drawing of an engine shows the relationships between its components by distributing them in the drawing area in a well-arranged and meaningful way.

The mapping ${\cal E}(P) \rightarrow P$ defined by $\mf{a} \mapsto \mf{a}_1$ is a strict homomorphism, and the mapping $P \rightarrow {\cal E}(P)$ with $x \mapsto \left( x, {\downarrow^{^{\!\!\!\circ}}} x, {\uparrow_{_{_{\!\!\!\circ}}}} x \right)$ is an embedding. We conclude $h_{{\cal E}(P)} = h_P$.

The following equations about the EV-systems of posets combined by order arithmetic follow directly from the definition. For every poset $P \in  \myfr{P} $ we have
\begin{align}
{\cal E}(P^d) & \simeq {\cal E}(P)^d, \label{EVsys_dual}
\end{align}
and for disjoint posets $P, Q \in  \myfr{P} $ we have
\begin{align}
{\cal E}(P+Q) = & \;\;\; {\cal E}(P) + {\cal E}(Q), \label{EVsys_dirsum} \\
{\cal E}( P \oplus Q ) = & \;\;\; \mysetdescr{ (x, D, U \cup V) }{(x,D,U) \in {\cal E}(P), V \subseteq Q} \label{EVsys_ordsum} \\
& \cup  \mysetdescr{ (x, D \cup E, U) }{(x,D,U) \in {\cal E}(Q), E \subseteq P}. \nonumber
\end{align}
Furthermore, ${\cal E}( P \times Q )$ is for any posets $P, Q \in  \myfr{P} $ the set of the $( (x,y), D, U )$ with
\begin{align}
(x,y) & \in P \times Q, \label{EVsys_prod} \\
D & \subseteq \left( ( {\downarrow \,}_P x ) \times ( {\downarrow \,}_Q y ) \right) \setminus \{ (x,y) \}, \nonumber \\
U & \subseteq \left( ( {\uparrow \,}_P x ) \times ( {\uparrow \,}_Q y ) \right) \setminus \{ (x,y) \}. \nonumber
\end{align}

Let $P, Q \in  \myfr{P} $ and $\xi \in {\cal H}(P,Q)$. We define a homomorphism $\alpha_{P,\xi} : P \rightarrow {\cal E}(Q)$ be setting for every $x \in P$
\begin{align*}
\myapxix{P}{\xi}{x} & \equiv \left( \xi(x), \xi( {\myodownarrow} \gxix ),
                          \xi( {\myouparrow} \gxix ) \right).
\end{align*}
If $P$ is fixed, we write $\myaxix{\xi}{x}$ instead of $\myapxix{P}{\xi}{x}$.

\begin{lemma}[{\cite[Proposition 2]{aCampo_toappear_1}}] \label{lemma_exix}
Let $R, S \in  \myfr{P} $, and let $\epsilon : {\cal E}(R) \rightarrow {\cal E}(S)$ be a strict homomorphism. For given $P \in  \myfr{P} $ and $\xi \in {\cal H}(P,R)$, we define for every $x \in P$
\begin{align*}
\eta(x) & \equiv \epsilon( \myaxix{\xi}{x} )_1,
\end{align*}
Then $\eta \in {\cal H}(P,S)$, and for every $x \in P$ we have
\begin{align*}
\mygxix{\eta}{x} & = \mygxix{\xi}{x}.
\end{align*}
\end{lemma}

The central definition is:

\begin{definition} \label{def_RS_scheme}
Let $R, S \in  \myfr{P} $. We call a mapping
\begin{align*}
\rho & \in \prod_{P \in  \myfr{P} _r} {\cal A}( {\cal H}(P,R), {\cal H}(P,S) ) \quad \mytext{(Cartesian product)}
\end{align*}
a {\em Hom-scheme from $R$ to $S$}. We call a Hom-scheme $\rho$ from $R$ to $S$
\begin{itemize}
\item {\em strong} iff the mapping $\rho_P : {\cal H}(P,R) \rightarrow {\cal H}(P,S)$ is one-to-one for every $P \in  \myfr{P} _r$;
\item a {\em G-scheme} iff for every $P \in  \myfr{P} _r, \xi \in {\cal H}(P,R), x \in P$
\begin{align*}
G_{\rho_P(\xi)}(x) & = \mygxix{\xi}{x};
\end{align*}
\item {\em image-controlled} or an {\em I-scheme}, iff for every $P, Q \in  \myfr{P} _r$, $\xi \in {\cal H}(P,R), \zeta \in {\cal H}(Q,R)$, $x \in P$, $y \in Q$
\begin{equation*}
\myapxix{P}{\xi}{x} \leq_+ \myapxix{Q}{\zeta}{y} \; \Rightarrow \;
\myapxix{P}{\rho_P(\xi)}{x} \leq_+ \myapxix{Q}{\rho_Q(\zeta)}{y},
\end{equation*}
where ``$<_+$'' on the left side implies ``$<_+$'' on the right side.
\end{itemize}
If $P$ is fixed, we write $\rho(\xi)$ instead of $\rho_P(\xi)$.\end{definition}
If a one-to-one homomorphism $\sigma : R \rightarrow S$ exists, we get a strong I-scheme from $R$ to $S$ by defining $\rho_P(\xi) \equiv \sigma \circ \xi$ for all $P \in  \myfr{P} $, $ \xi \in {\cal H}(P,R)$. For an image-controlled Hom-scheme $\rho$, we have due to the antisymmetry of $\leq_+$
\begin{equation} \label{eq_imagebased_rhowert}
\myapxix{P}{\xi}{x} = \myapxix{Q}{\zeta}{y}
\; \Rightarrow \;
\myapxix{P}{\rho_P(\xi)}{x} = \myapxix{Q}{\rho_Q(\zeta)}{y}.
\end{equation}

An I-scheme is always a G-scheme \cite[Theorem 4]{aCampo_toappear_1}. We write
\begin{equation*}
R \sqsubseteq S \; / \; R \sqsubseteq_G S \; / \; R \sqsubseteq_I S
\end{equation*}
iff a strong Hom-scheme / a strong G-scheme / a strong I-scheme exists from $R$ to $S$. The three relations $\sqsubseteq$, $\sqsubseteq_G$, and $\sqsubseteq_I$ define partial orders on $ \myfr{P} _r$ \cite[Theorems 2 and 3]{aCampo_toappear_1}.

The main results of \cite{aCampo_toappear_1,aCampo_toappear_2} about strong I-schemes and strong G-schemes are:

\begin{theorem}[{\cite[Theorems 5 and 7]{aCampo_toappear_1}}] \label{theo_strongI_equiv_eps}
For posets $R, S \in  \myfr{P} $, we have $ R \sqsubseteq_I S$ iff there exists a one-to-one homomorphism $\epsilon : {\cal E}(R) \rightarrow {\cal E}(S)$ with
\begin{equation} \label{eq_cond_univ_aexix}
\myapexix{P}{\xi}{x} \quad = \quad \epsilon( \myapxix{P}{\xi}{x}  ).
\end{equation}
for every $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$, where
\begin{equation*}
\eta(\xi)(x) \; \equiv \; \myepxix{P}{\xi}{x} \; \equiv \; \epsilon( \myapxix{P}{\xi}{x} )_1
\end{equation*}
for every $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$. In particular, $\eta$ is a strong I-scheme from $R$ to $S$.
\end{theorem}

\begin{theorem}[{\cite[Theorem 1]{aCampo_toappear_2}}] \label{theo_GschemeOnStrict}
Let $R, S \in  \myfr{P} $. Equivalent are
\begin{align*}
R & \sqsubseteq_G S; \\
\# {\cal S}(P,R) & \leq \# {\cal S}(P,S) \; \; \mytext{for all} \; P \in  \myfr{P} ;  \\
\# {\cal S}(P,R) & \leq \# {\cal S}(P,S) \; \; \mytext{for all connected} \; P \in  \myfr{P} .
\end{align*}
\end{theorem}
The following proposition is not contained in \cite{aCampo_toappear_1,aCampo_toappear_2}. It shows that a mapping behaving like a (strong) Hom-scheme, G-scheme, or I-scheme on the set of the connected posets in $ \myfr{P} _r$ can be extended to a (strong) Hom-scheme, G-scheme, or I-scheme (on the total set $ \myfr{P} _r$):

\begin{proposition} \label{prop_zshP_reicht}
Let
\begin{align*}
\tau & \in \prod_{\stackrel{Q \in  \myfr{P} _r}{Q \mytext{connected}}} {\cal A}( {\cal H}(Q,R), {\cal H}(Q,S) ).
\end{align*}
For $P \in  \myfr{P} _r$ with connectivity components $K_j$, $j \in \myNk{J}$, $J \in \mathbb{N}$, we define for every $\xi \in {\cal H}(P,R)$
\begin{align*}
\rho_P( \xi ) & \equiv \bigcup_{j=1}^J \tau_{K_j}( \xi \vert_{K_j} ).
\end{align*}
Then $\rho$ is a Hom-scheme, and we have:
\begin{itemize}
\item $\rho$ is strong iff $\tau_Q : {\cal H}( Q, R ) \rightarrow {\cal H}(Q,S)$ is one-to-one for every connected $Q \in  \myfr{P} _r$;
\item $\rho$ is a G-scheme iff $\mygxix{\tau_Q(\zeta)}{y} = \mygxix{\zeta}{y}$ for every connected $Q \in  \myfr{P} _r$, $\zeta \in {\cal H}(Q,R)$, and $y \in Q$;
\item $\rho$ is an I-scheme iff for every connected $Q, Q^* \in  \myfr{P} _r$, $\zeta \in {\cal H}(Q,R), \zeta^* \in {\cal H}(Q^*,R)$, $y \in Q$, $y^* \in Q^*$
\begin{equation} \label{eq_imagebased_in_Th}
\myapxix{Q}{\zeta}{y} \leq_+ \myapxix{Q^*}{\zeta^*}{y^*} \; \Rightarrow \;
\myapxix{Q}{\tau_Q(\zeta)}{y} \leq_+ \myapxix{Q^*}{\tau_{Q^*}(\zeta^*)}{y^*},
\end{equation}
where ``$<_+$'' on the left side implies ``$<_+$'' on the right side.
\end{itemize}
\end{proposition}
\begin{proof} In all three proofs we have to show ``$\Leftarrow$'' only, because if $\tau$ fails for one of the three properties, then obviously also $\rho$ does.

It is clear that $\rho$ is a Hom-scheme. In the proofs of the first two statements, $P \in  \myfr{P} _r$ with connectivity components $K_j$, $j \in \myNk{J}$, $J \in \mathbb{N}$, is fixed.

Let $\tau_Q$ be one-to-one for every connected $Q \in  \myfr{P} _r$, and let $\xi, \xi' \in {\cal H}(P,R)$ with $\xi \not= \xi'$. There exists an $x \in P$ with $\xi(x) \not= \xi'(x)$ and a $j \in \myNk{J}$ with $x \in K_j$. We conclude $\xi \vert_{K_j} \not= \xi' \vert_{K_j}$, thus $\tau( \xi \vert_{K_j} ) \not= \tau( \xi' \vert_{K_j} )$, and $\rho(\xi) \not= \rho(\xi')$ is proven.

Assume $\mygxix{\tau_Q(\zeta)}{y} = \mygxix{\zeta}{y}$ for all connected $Q \in  \myfr{P} _r$, $\zeta \in {\cal H}(Q,R)$, and $y \in Q$. Let $\xi \in {\cal H}(P,R)$ and $x \in K_j$. We have $\mygxix{\rho(\xi)}{x} = \mygxix{\xi}{x} \subseteq K_j$, and applying \eqref{gxix_postrestr_unten} twice yields
\begin{align*}
\mygxix{\xi}{x} & = \mygxix{\xi \vert_{K_j}}{x}
= \mygxix{\tau( \xi \vert_{K_j})}{x}
= \mygxix{\rho( \xi )\vert_{K_j}}{x} = \mygxix{\rho(\xi)}{x}.
\end{align*}

Now assume that \eqref{eq_imagebased_in_Th} holds for all connected posets $Q, Q^* \in  \myfr{P} _r$, $\zeta \in {\cal H}(Q,R), \zeta^* \in {\cal H}(Q^*,R)$, $y \in Q$, $y^* \in Q^*$. Let $P, P^* \in  \myfr{P} _r$, $\xi \in {\cal H}(P,R), \xi^* \in {\cal H}(P^*,R)$, $x \in P$, $x^* \in P^*$ with $\myapxix{P}{\xi}{x} \leq_+ \myapxix{P^*}{\xi^*}{x^*}$. The connectivity components of $P$ are $K_j$, $j \in \myNk{J}$, $J \in \mathbb{N}$, and the connectivity components of $P^*$ are $K^*_{j^*}$, $j^* \in \myNk{J^*}$, $J^* \in \mathbb{N}$. Let $j_0 \in \myNk{J}$ with $x \in K_{j_0}$ and $j^*_0 \in \myNk{J^*}$ with $x^* \in K^*_{j^*_0}$.

Due to $ \mygxix{\xi}{x} = \mygxix{\xi \vert_{K_{j_0}}}{x} \subseteq K_{j_0}$, we get
\begin{align*}
\myapxix{P}{\xi}{x}_2 & = \; \xi( {\myodownarrow} \gxix ) \;
\stackrel{\eqref{gxix_postrestr_unten}}{=}
\; \xi( \downarrow^{^{\!\!\!\circ}} \mygxix{\xi \vert_{K_{j_0}}}{x}  )
\; = \; \xi_{K_{j_0}}( \downarrow^{^{\!\!\!\circ}} \mygxix{\xi \vert_{K_{j_0}}}{x} ) \\
& = \; \myapxix{K_{j_0}}{\xi \vert_{K_{j_0}}}{x}_2
\end{align*}
and, in the same way,
$\myapxix{P}{\xi}{x}_3 = \myapxix{K_{j_0}}{\xi \vert_{K_{j_0}}}{x}_3$. Because $\xi( x ) = \xi \vert_{K_{j_0}}(x)$ is trivial, we have
$\myapxix{P}{\xi}{x} = \myapxix{K_{j_0}}{\xi \vert_{K_{j_0}}}{x}$. In the same way we see
\begin{align*}
\myapxix{P^*}{\xi^*}{x^*} & = \myapxix{K^*_{j^*_0}}{\xi^* \vert_{K^*_{j^*_0}}}{x^*}, \\
\myapxix{P}{\rho(\xi)}{x} & = \myapxix{K_{j_0}}{\tau( \xi \vert_{K_{j_0}})}{x}, \\
\myapxix{P^*}{\rho(\xi^*)}{x^*} & = \myapxix{K^*_{j^*_0}}{\tau(\xi^* \vert_{K^*_{j^*_0}})}{x^*}.
\end{align*}
Now we get
\begin{align*}
\myapxix{K_{j_0}}{\xi \vert_{K_{j_0}}}{x} & \; = \; \myapxix{P}{\xi}{x} \; \leq_+ \; \myapxix{P^*}{\xi^*}{x^*} \; = \; \myapxix{K^*_{j^*_0}}{\xi^* \vert_{K^*_{j^*_0}}}{x^*}
\end{align*}
and \eqref{eq_imagebased_in_Th} delivers
\begin{align*}
\myapxix{P}{\rho(\xi)}{x} & \; = \; \myapxix{K_{j_0}}{\tau( \xi \vert_{K_{j_0}})}{x} \; \leq_+ \; \myapxix{K^*_{j^*_0}}{\tau(\xi^* \vert_{K^*_{j^*_0}})}{x^*} \; = \; \myapxix{P^*}{\rho(\xi^*)}{x^*}
\end{align*}
with ``$<_+$'' if $\myapxix{P}{\xi}{x} <_+ \myapxix{P^*}{\xi^*}{x^*}$.

\end{proof}

\section{Compatibility with order arithmetic} \label{sec_calcrules}

In this section, we examine to which extent the relations $\sqsubseteq, \sqsubseteq_G$, and $\sqsubseteq_I$ are compatible with order arithmetic. The simple equation \eqref{EVsys_dual} for ${\cal E}(P^d)$ lets us expect that there will be a simple relation between the different types of strong Hom-schemes from $R$ to $S$ and from $R^d$ to $S^d$. Similarly, looking at \eqref{EVsys_dirsum}, the direct sum of posets should also be easily manageable in the world of strong Hom-schemes. The formulas \eqref{EVsys_ordsum} and \eqref{EVsys_prod} for ${\cal E}(P \oplus Q)$ and ${\cal E}(P \times Q)$ refer to ${\cal E}(P)$ and ${\cal E}(Q)$ in a more involved way, and therefore, the situation will be more complicated for ordinal sums and products of posets.

\begin{proposition} \label{calc_dual}
Let $R, S \in  \myfr{P} $. Then
\begin{align*}
R \sqsubseteq S & \Leftrightarrow R^d \sqsubseteq S^d, \\
R \sqsubseteq_G S & \Leftrightarrow R^d \sqsubseteq_G S^d, \\
R \sqsubseteq_I S & \Leftrightarrow R^d \sqsubseteq_I S^d, \\
R \sqsubseteq R^d & \Leftrightarrow R \simeq R^d.
\end{align*}
\end{proposition}
\begin{proof} Let $\rho$ be a strong Hom-scheme from $R$ to $S$. Because of ${\cal H}(P,Q) = {\cal H}(P^d, Q^d)$ for every $P, Q \in  \myfr{P} $, we get a strong Hom-scheme $\rho^d$ from $R^d$ to $S^d$ by setting $\rho^d_P \equiv \rho_{P^d}$ for all $P \in  \myfr{P} $. Obviously, $\rho^d$ is a G-scheme (an I-scheme) iff $\rho$ is. The last proposition follows from the first one due to the antisymmetry of $\sqsubseteq$.

\end{proof}

In order to avoid repetitions, we agree that the carriers of posets involved in direct or ordinal sums are always disjoint. Additionally, we agree on that the index $j$ is always an element of $\myNk{2}$.

\begin{proposition} \label{calc_dirsum}
Let $R^1, R^2, S^1, S^2 \in  \myfr{P} $. Then
\begin{align*}
R^1 \sqsubseteq S^1 \mytext{and} R^2 \sqsubseteq S^2 & \; \Rightarrow \; R^1 + R^2 \sqsubseteq S^1 + S^2, \\
R^1 \sqsubseteq_G S^1 \mytext{and} R^2 \sqsubseteq_G S^2 & \; \Rightarrow \; R^1 + R^2 \sqsubseteq_G S^1 + S^2, \\
R^1 \sqsubseteq_I S^1 \mytext{and} R^2 \sqsubseteq_I S^2 & \; \Rightarrow \; R^1 + R^2 \sqsubseteq_I S^1 + S^2.
\end{align*}
\end{proposition}
\begin{proof} Let $\rho^j$ be a strong Hom-scheme / G-scheme / I-scheme from $R^j$ to $S^j$. For every connected $P \in  \myfr{P} $, we have ${\cal H}(P,R^1+R^2) \simeq {\cal H}(P,R^1) + {\cal H}(P,R^2)$ and ${\cal H}(P,S^1+S^2) \simeq {\cal H}(P,S^1) + {\cal H}(P,S^2)$. We define $\rho^+_P \equiv \rho^1_P \cup \rho^2_P$ for every connected $P \in  \myfr{P} $, and all statements follow with Proposition \ref{prop_zshP_reicht}. (Observe $\rho_P^+(\xi) = \rho_P^j(\xi)$, $\mygxix{\rho^+(\xi)}{x} = \mygxix{\rho^j(\xi)}{x} $, and $\myaxix{\rho^+(\xi)}{x} = \myaxix{\rho^j(\xi)}{x}$ for every $\xi \in {\cal H}(P,R^j), x \in P$, $P \in  \myfr{P} _r$ connected.)

\end{proof}

\begin{proposition} \label{calc_ordsum}
Let $R^1, R^2, S^1, S^2 \in  \myfr{P} $. Then
\begin{align*}
R^1 \sqsubseteq S^1 \mytext{and} R^2 \sqsubseteq S^2 & \; \Rightarrow \; R^1 \oplus R^2 \sqsubseteq S^1 \oplus S^2, \\
R^1 \sqsubseteq_G S^1 \mytext{and} R^2 \sqsubseteq_G S^2 & \; \Rightarrow \; R^1 \oplus R^2 \sqsubseteq_G S^1 \oplus S^2.
\end{align*}
\end{proposition}
\begin{proof} For posets $A, B$, the relation $A \sqsubseteq B$ means $\# {\cal H}(P,A) \leq \# {\cal H}(P,B)$ for all $P \in  \myfr{P} $, and according to Theorem \ref{theo_GschemeOnStrict}, the relation $A \sqsubseteq_G B$ is equivalent to $\# {\cal S}(P,A) \leq \# {\cal S}(P,B)$ for all $P \in  \myfr{P} $. The statements are now direct consequences of
\begin{align*}
{\cal H}(P, A \oplus B) \simeq \sum_{U \in {\cal U}(P)} {\cal H}(X \setminus U, A) \times {\cal H}(U, B), \\
{\cal S}(P, A \oplus B) \simeq \sum_{U \in {\cal U}(P)} {\cal S}(X \setminus U, A) \times {\cal S}(U, B),
\end{align*}
for all finite posets $P, A, B$, where ${\cal U}(P)$ is the set of the upsets of $P$ and where $X \setminus U$ and $U$ are equipped with the partial order induced by $P$.

\end{proof}

\begin{proposition} \label{calc_prod}
Let $R^1, R^2, S^1, S^2 \in  \myfr{P} $. Then
\begin{align*}
R^1 \sqsubseteq S^1 \mytext{and} R^2 \sqsubseteq S^2 & \; \Rightarrow \; R^1 \times R^2 \sqsubseteq S^1 \times S^2, \\
R^1 \sqsubseteq_G S^1 \mytext{and} R^2 \sqsubseteq_G S^2 & \; \Rightarrow \; R^1 \times R^2 \sqsubseteq_G S^1 \times S^2.\end{align*}
\end{proposition}
\begin{proof} The first statement is a direct consequence of ${\cal H}(P, A \times B) \simeq {\cal H}(P, A) \times {\cal H}(P,B)$ for all finite posets $P, A, B$. For a constructive proof (which we need for the second statement), take strong Hom-schemes $\rho^j$ from $R^j$ to $S^j$. For every $P \in  \myfr{P} _r$ and every $\xi = ( \xi^1, \xi^2 ) \in {\cal H}( P, R^1 \times R^2 )$ (with $ \xi^j \in {\cal H}(P , R^j )$), we define $\rho^\times_P((\xi^1, \xi^2)) \equiv ( \rho^1_P(\xi^1), \rho^2_P(\xi^2))$. Then $\rho^\times$ is a strong Hom-scheme from $R^1 \times R^2$ to $S^1 \times S^2$.

Now let $\rho^1$ and $\rho^2$ be strong G-schemes. For every poset $P \in  \myfr{P} _r$, $\xi = ( \xi^1, \xi^2 ) \in {\cal H}(P,R^1 \times R^2)$, $x \in P$, we get by applying Lemma \ref{lemma_G_prod} twice
\begin{equation*}
\mygxix{\rho^\times(\xi)}{x} \; = \; \gamma_{ \mygxix{\rho^1(\xi^1)}{x} \cap \mygxix{\rho^2(\xi^2)}{x} }(x)
\; = \; \gamma_{ \mygxix{\xi^1}{x} \cap \mygxix{\xi^2}{x} }(x)
\; = \; \mygxix{\xi}{x}.
\end{equation*}

\end{proof}

\begin{proposition} \label{calc_HomSets}
Let $R, S\in  \myfr{P} $. Then
\begin{equation} \label{HomSets_1}
R \sqsubseteq S \quad \Rightarrow \quad {\cal H}(Q,R) \sqsubseteq_G {\cal H}(Q,S) \; \mytext{for every} \; Q \in  \myfr{P} _r.
\end{equation}
Moreover,
\begin{equation} \label{HomSets_2}
\forall \; Q \in  \myfr{P} _r \; : \; {\cal H}(Q,R) \sqsubseteq \; \; {\cal H}(Q,S) \quad \Rightarrow \quad R \sqsubseteq S,
\end{equation}
and thus
\begin{align} \label{HomSets_3}
& \forall \; Q \in  \myfr{P} _r \; : \; {\cal H}(Q,R) \sqsubseteq \; \; {\cal H}(Q,S) \\
\Leftrightarrow \quad & \forall \; Q \in  \myfr{P} _r \; : \; {\cal H}(Q,R) \sqsubseteq_G {\cal H}(Q,S). \nonumber
\end{align}
Finally,
\begin{equation} \label{HomSets_4}
\forall \; Q \in  \myfr{P} _r \; : \; {\cal H}(Q,R) \sqsubseteq_I {\cal H}(Q,S) \quad
\Rightarrow \quad R \sqsubseteq_I S.
\end{equation}

\end{proposition}
\begin{proof} \eqref{HomSets_1}: Let $\rho$ be a strong Hom-scheme from $R$ to $S$, and let $Q \in  \myfr{P} _r$ be fixed. We define for every $P \in  \myfr{P} _r$ and every $\xi \in {\cal H}( P, {\cal H}(Q,R))$
\begin{align*}
\tau_P(\xi) & \equiv \rho_Q \circ \xi.
\end{align*}
Then $\tau_P : {\cal H}( P, {\cal H}(Q,R)) \rightarrow {\cal H}( P, {\cal H}(Q,S))$ is a well-defined mapping, and because $\rho_Q : {\cal H}(Q,R) \rightarrow {\cal H}(Q,S)$ is a one-to-one mapping, we have $\mygxix{\tau_P(\xi)}{x} = \mygxix{\xi}{x}$ for all $x \in P$. $\tau$ is thus a G-scheme from ${\cal H}(Q,R)$ to ${\cal H}(Q,S)$.

Let $\xi, \zeta \in {\cal H}( P, {\cal H}(Q,R))$ with $\tau_P(\xi) = \tau_P(\zeta)$. Then $\rho_Q( \xi(x) ) = \tau_P(\xi)(x) = \tau_P(\zeta)(x) = \rho_Q( \zeta(x) )$ for all $x \in P$. Because $\rho_Q$ is one-to-one, we get $\xi(x) = \zeta(x)$ for all $x \in P$, hence $\xi = \zeta$, and $\tau$ is strong.

\eqref{HomSets_2}: In the case of ${\cal H}(Q,R) \sqsubseteq {\cal H}(Q,S) $ for all $Q \in  \myfr{P} _r$, we have for all $ P \in  \myfr{P} _r$: $ \# {\cal H}(P,R) = \# {\cal H}(A_1, {\cal H}(P,R)) \leq \# {\cal H}( A_1, {\cal H}(P,S) )$ $ = \# {\cal H}(P,S)$, thus $ R \sqsubseteq S$.

\eqref{HomSets_3}: ${\cal H}(Q,R) \sqsubseteq {\cal H}(Q,S) $ for all $Q \in  \myfr{P} _r$ yields $R \sqsubseteq S$ according to \eqref{HomSets_2}, and \eqref{HomSets_1} delivers ${\cal H}(Q,R) \sqsubseteq_G {\cal H}(Q,S) $ for all $Q \in  \myfr{P} _r$. The direction ``$\Leftarrow$'' is trivial.

\eqref{HomSets_4}: $ R \simeq {\cal H}(A_1, R) \sqsubseteq_I {\cal H}(A_1, S) \simeq S$.

\end{proof}

\section{The ansatz for the proof of cancellation rules} \label{sec_ansatz_cancel}

In the following sections, we deal with cancellation rules for strong Hom-schemes: For $ \odot \in \{ +, \oplus, \times \}$ and ${\preceq} \in \{ \sqsubseteq, \sqsubseteq_G, \sqsubseteq_I  \}$, we have posets $Q, R, S \in  \myfr{P} $ with $ Q \odot R \preceq Q \odot S$. Does this imply $R \preceq S$? For the direct sum of posets (Section \ref{sec_cancel_dirsum}), the answer is ``Yes'' with no ifs and buts, but for ordinal sums and products (Sections \ref{sec_cancel_ordsum} and \ref{sec_cancel_product}), we need additional assumptions in most cases.

All together we prove eight cancellation rules (as already announced in the introduction, we did not succeed in proving a cancellation rule for $ Q \oplus R \sqsubseteq Q \oplus S$). In three of the proofs, the inequality between the cardinalities of the respective homomorphism sets is directly proven. The remaining five proofs use a common ansatz. Obviously, we can embed ${\cal H}(P,R)$ into ${\cal H}(P, Q \odot R)$ and ${\cal H}(P,S)$ into ${\cal H}(P, Q \odot S)$ for all $ \odot \in \{ +, \oplus, \times \}$, and \eqref{EVsys_dirsum}, \eqref{EVsys_ordsum}, and \eqref{EVsys_prod} show that also the EV-systems ${\cal E}(R)$ and ${\cal E}(S)$ can be embedded into ${\cal E}(Q \odot R )$ and ${\cal E}(Q \odot S )$, respectively, for all $ \odot \in \{ +, \oplus, \times \}$. We therefore choose a suitable embedding of ${\cal H}(P,R)$ into ${\cal H}(P, Q \odot R)$ and drive it into an embedding of ${\cal H}(P,S)$ into ${\cal H}(P, Q \odot S)$ by applying the strong Hom-scheme / G-scheme / I-scheme $\rho$ from $Q \odot R$ to $Q \odot S$. Similarly, in the case of strong I-schemes, we choose an embedding of ${\cal E}(R)$ into ${\cal E}(Q \odot R )$ and push it into an embedding of ${\cal E}(S)$ into ${\cal E}(Q \odot S)$ by means of the one-to-one homomorphism $\epsilon$ from ${\cal E}(Q \odot R )$ to ${\cal E}( Q \odot S)$ described in Theorem \ref{theo_strongI_equiv_eps}.

In the case of $\odot = \oplus$, the pushing process succeeds for strong I-schemes with a single application of $\epsilon$ on the embedding of ${\cal E}(R)$, and also for strong G-schemes, it is enough to apply $\rho$ once. In the case of $ \odot \in \{ +, \times \}$, we apply $\rho$ and $\epsilon$ repeatedly within iterative processes, and we show that after a finite number of iteration steps the embedding of ${\cal H}(P,R)$ arrives at an embedding of ${\cal H}(P,S)$ (the embedding of ${\cal E}(R)$ arrives at an embedding of ${\cal E}(S)$). An interesting aspect is that for both $ \odot \in \{ +, \times \}$, the iteration processes itself rely on pure set theory without order theoretical aspects. Order theory and the theory of Hom-schemes are applied on the objects the iterations are stepping through and terminating with, in showing that they provide the appropriate order theoretical properties and relations.

\section{Cancellation rules for direct sums} \label{sec_cancel_dirsum}

\begin{theorem} \label{theo_cancel_dirsum_1}
Let $Q, R, S \in  \myfr{P} $ with $Q \cap R = \emptyset$ and $Q \cap S = \emptyset$. Then
\begin{align*}
Q + R \sqsubseteq \; \; Q + S & \quad \Rightarrow \quad R \sqsubseteq S, \\
Q + R \sqsubseteq_G Q + S & \quad \Rightarrow \quad R \sqsubseteq_G S.
\end{align*}
\end{theorem}
\begin{proof} Let $P \in  \myfr{P} $ be a connected poset. For all disjoint posets $A, B \in  \myfr{P} $, we have
\begin{align*}
{\cal H}(P, A+B) & \simeq {\cal H}(P,A) + {\cal H}(P,B), \\
{\cal S}(P, A+B) & \simeq {\cal S}(P,A) + {\cal S}(P,B).
\end{align*}
$ \# {\cal H}(P, Q+R) \leq {\cal H}(P, Q+S) $ yields thus $ \# {\cal H}( P, R ) \leq \# {\cal H}( P, S )$, and with Proposition \ref{prop_zshP_reicht} we conclude $ R \sqsubseteq S$. In the same way, we get $ R \sqsubseteq_G S$ by using Theorem \ref{theo_GschemeOnStrict}.

\end{proof}

For the proof of the cancellation rule for $\sqsubseteq_I$, we need a lemma belonging to set theory:

\begin{lemma} \label{lemma_injAbb}
Let $A, B, C$ be non-empty, disjoint sets, $A$ finite, and let the mapping $f : A \cup B \rightarrow A \cup C$ be one-to-one. For every $b \in B$, there exists an integer $n(b) \in \mathbb{N}$ with
\begin{align*}
f^{n(b)}(b) & \in C, \\
f^\nu(b) & \in A \; \mytext{for} \; 1 \leq \nu \leq n(b) - 1,
\end{align*}
and the mapping
\begin{align*}
F : B & \rightarrow C, \\
b & \mapsto f^{n(b)}(b)
\end{align*}
is one-to-one.
\end{lemma}
\begin{proof} Assume $f^\nu(b) \in A$ for all $\nu \in \mathbb{N}$. Because $A$ is finite, there exist $i, j \in \mathbb{N}$ with $i > j$ and $f^i(b) = f^j(b)$. Because $f$ is one-to-one, we get $b = f^{i-j}(b) \in A$, a contradiction to $A \cap B = \emptyset$.

Let $x, y \in B$ with $F(x) = F(y)$, thus $f^{n(x)}(x) = f^{n(y)}(y)$. Assume $n(x) \geq n(y)$ without loss of generality. Because $f$ is one-to-one, we get $f^{n(x) - n(y)}(x) = y \in B$, and we conclude $n(x) - n(y) = 0$, thus $x = y$.

\end{proof}

\begin{theorem} \label{theo_cancel_dirsum_2}
Let $Q, R, S \in  \myfr{P} $, with $Q \cap R = \emptyset$ and $Q \cap S = \emptyset$. Then
\begin{align*}
Q + R \sqsubseteq_I Q + S & \quad \Rightarrow \quad R \sqsubseteq_I S.
\end{align*}
\end{theorem}
\begin{proof} According to Theorem \ref{theo_strongI_equiv_eps} and \eqref{EVsys_dirsum}, there exists a one-to-one homomorphism $\epsilon : {\cal E}(Q) + {\cal E}(R) \rightarrow {\cal E}(Q) + {\cal E}(S)$ with $\myapexix{P}{\xi}{x} = \epsilon( \myapxix{P}{\xi}{x}  )$ for every $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$, where $\eta(\xi)(x) \; \equiv \epsilon( \myapxix{P}{\xi}{x} )_1.$ Due to Lemma \ref{lemma_injAbb}, there exists for every $\mf{a} \in {\cal E}(R)$ an integer $n(\mf{a}) \in \mathbb{N}$ with $\epsilon^{n(\mf{a})}(\mf{a}) \in {\cal E}(S)$ and
$\epsilon^\nu(\mf{a}) \in {\cal E}(Q)$ for all $\nu \in \myNk{n(\mf{a}) - 1}$; additionally, the mapping
\begin{align*}
E : {\cal E}(R) & \rightarrow {\cal E}(S), \\
\mf{a} & \mapsto \epsilon^{n(\mf{a})}(\mf{a})
\end{align*}
is one-to-one. Let $\mf{a}, \mf{b} \in {\cal E}(R)$ with $\mf{a} \leq_+ \mf{b}$. Then $\mf{a}$ and $\mf{b}$ belong to the same connectivity component of ${\cal E}(R)$, and with $j \equiv \min\{ n(\mf{a}), n(\mf{b}) \}$, the points $\epsilon^j( \mf{a} ) \leq_+ \epsilon^j( \mf{b} )$ belong to the same connectivity component of ${\cal E}(Q) + {\cal E}(S)$. According to the definition of $j$, at least one of the points $\epsilon^j( \mf{a} )$ and $\epsilon^j( \mf{b} )$ belongs to ${\cal E}(S)$, thus $\epsilon^j( \mf{a} ), \epsilon^j( \mf{b} ) \in {\cal E}(S)$, hence $ n(\mf{a}) =  n(\mf{b})$. We conclude $E( \mf{a} ) = \epsilon^{n(\mf{a})}( \mf{a} ) \leq_+ \epsilon^{n(\mf{a})}( \mf{b} ) = \epsilon^{n(\mf{b})}( \mf{b} ) = E(\mf{b})$, and $E$ is a one-to-one homomorphism.

Let $P \in  \myfr{P} _r, \xi \in {\cal H}(P, Q + R)$, $x \in P$. By induction, it is easily seen that $\epsilon^\nu( \myaxix{\xi}{x} ) = \myaxix{ \eta^\nu(\xi) }{x}$ for all $\nu \in \myNk{ n(\myaxix{\xi}{x}) }$. (Be aware that for $\nu \in \myNk{ n(\myaxix{\xi}{x}) - 1} $, we have $\epsilon^\nu( \myaxix{\xi}{x} ) \in {\cal E}(Q) \subseteq {\cal E}(Q) + {\cal E}(R)$, hence $\eta^\nu(\xi) \in {\cal H}(P, Q + R)$; the mapping $\eta^{\nu+1}(\xi)$ is thus well-defined.) Now we conclude $E( \myaxix{\xi}{x} )_1 = \eta^{n(\myaxix{\xi}{x})}(\xi)(x)$ and
\begin{equation*}
E( \myaxix{\xi}{x} ) \; = \; \epsilon^{n(\myaxix{\xi}{x})}(\myaxix{\xi}{x}) \; = \; \myaxix{ \eta^{n(\myaxix{\xi}{x})}(\xi) }{x},
\end{equation*}
and $E$ fulfills the requirements of Theorem \ref{theo_strongI_equiv_eps}.

\end{proof}

\section{Cancellation rules for ordinal sums} \label{sec_cancel_ordsum}

As already announced in the introduction, we did not succeed in proving a cancellation rule for $ Q \oplus R \sqsubseteq Q \oplus S$.

\begin{theorem} \label{theo_cancel_ordsum_G}
Let $Q, R, S \in  \myfr{P} $ with $Q \cap R = \emptyset$ and $Q \cap S = \emptyset$. Then
\begin{align*}
Q \oplus R \sqsubseteq_G Q \oplus S & \quad \Rightarrow \quad R \sqsubseteq_G S.
\end{align*}
\end{theorem}
\begin{proof} For every $ P \in  \myfr{P}  $, we define for every $\xi \in {\cal H}(P,R)$
\begin{align*}
\Gamma_{P,R}(\xi) & \; \equiv \; \mysetdescr{ \zeta \in {\cal H}(P,R) }{ \mygxix{\zeta}{x} = \mygxix{\xi}{x} \; \mytext{for all} \; x \in P}, \\
\Gamma_{P,S}(\xi) & \; \equiv \; \mysetdescr{ \zeta \in {\cal H}(P,S) }{ \mygxix{\zeta}{x} = \mygxix{\xi}{x} \; \mytext{for all} \; x \in P}.
\end{align*}
As proven in \cite[Lemma 4]{aCampo_toappear_2}, $R \sqsubseteq_G S$ is equivalent to
\begin{align} \label{ungl_GRxi_GSxi}
 \# \Gamma_{P,R}(\xi) & \leq \# \Gamma_{P,S}(\xi) \quad \mytext{for all} \; \xi \in {\cal H}(P,R), P \in  \myfr{P} .
\end{align}

Let $\rho$ be a strong G-scheme from $Q \oplus R $ to $Q \oplus S$, and let $P \in  \myfr{P} _r$ be fixed. For every $k \in \mathbb{N}$, let $kP$ be the direct sum of $k$ disjoint isomorphic copies of $P$. We identify ${\cal H}( kP, R)$ with ${\cal H}(P, R)^k$ and ${\cal H}( kP, S)$ with ${\cal H}(P, S)^k$. Without loss of generality we can assume $Q \cap kP = \emptyset$ for every $k \in \mathbb{N}$. (Otherwise, we replace $Q$ by a suitable isomorphic copy.)

Let $\xi \in {\cal H}(P,R)$ be fixed. We define for every $k \in \mathbb{N}$
\begin{align*}
M_k & \equiv \mysetdescr{ \zeta \cup \chi }{ \zeta \in {\cal S}(Q,Q), \chi \in \Gamma_{P,R}(\xi)^k }
\end{align*}
$M_k$ is isomorphic to a subset of $ {\cal H}(Q,Q) \times {\cal H}( kP, R )$ which is again isomorphic to a subset of ${\cal H}(Q \oplus kP, Q \oplus R )$. $\rho_k( \theta ) \equiv \rho_{Q \oplus kP}(\theta)$ is thus well-defined for every $\theta \in M_k$.

Let $\theta = \zeta \cup \chi \in M_k$ with $\zeta \in {\cal S}(Q,Q)$, $\chi \in \Gamma_{P,R}( \xi )^k$. We have $\theta(Q) \subseteq Q$ and $\theta( kP ) \subseteq R$ with $Q \cap R = \emptyset$, which yields $\mygxix{\theta}{y} = \mygxix{\zeta}{y}$ for $y \in Q$. Therefore, for $y \in Q$, we get $\mygxix{\rho_k(\theta)}{y} = \mygxix{\theta}{y} = \{ y \}$, because $\zeta$ is strict. We conclude $ \rho_k(\theta) \vert_Q \in {\cal S}(Q, Q \oplus S)$.

Now we show $\rho_k(\theta)( k P ) \subseteq S$. Let $y_1 < \ldots < y_L$ be a chain of maximal length in $Q$. Due to the strictness of $\rho_k(\theta) \vert_Q$, we have $\rho_k(\theta)(y_1) < \ldots < \rho_k(\theta)(y_L)$. Assume $\rho_k(\theta)(x) \in Q$ for an $x \in kP$. Due to $y_\ell < x$ in $Q \oplus kP$ for all $\ell \in \myNk{L}$, we have $\rho_k(\theta)(y_\ell) \leq \rho_k(\theta)(x)$, and, in particular, $\rho_k(\theta)(y_\ell) \in Q$ for all $\ell \in \myNk{L}$. We conclude $\rho_k(\theta)(y_L) = \rho_k(\theta)(x)$, because in the case of ``$<$'' we get a chain of length $L+1$ in $Q$. But in the case $\rho_k(\theta)(y_L) = \rho_k(\theta)(x)$, we have $x \in \mygxix{\rho_k(\theta)}{y_L} = \{ y_L \}$ in contradiction to $Q \cap kP = \emptyset$. Thus, $\rho_k(\theta)( k P ) \subseteq S$. In particular, $\rho_k(\theta) \vert_{kP}^S$ is well-defined.

Let $x \in kP$. Due to $\theta(Q) \subseteq Q$ and $\theta( kP ) \subseteq R$ with $Q \cap R = \emptyset$, we have $ \mygxix{\theta}{x} \subseteq kP $, hence $ \mygxix{\rho_k(\theta)}{x} \subseteq kP $, too. Now we get
\begin{align*}
\mygxix{\rho_k(\theta) \vert_{kP}^S}{x} &
\stackrel{(\ref{gxix_postrestr_oben},\ref{gxix_postrestr_unten})}{=}
\mygxix{\rho_k(\theta)}{x} = \mygxix{\theta}{x}
\stackrel{\eqref{gxix_postrestr_unten}}{=}
\mygxix{\theta \vert_{kP}}{x}
= \mygxix{\chi}{x}
= \mygxix{\xi}{x},
\end{align*}
thus $\rho_k(\theta) \vert_{kP}^S \in \Gamma_{P,S}( \xi )^k$. All together, we can write $\rho_k(\theta) = \zeta' \cup \chi'$ with $\zeta' \in {\cal S}( Q, Q \oplus S)$ and $\chi' \in \Gamma_{P,S}( \xi )^k$. Accordingly, we can regard $\rho_k( M_k )$ as a subset of ${\cal S}( Q, Q \oplus S) \times \Gamma_{P,S}( \xi )^k$. Now we get
\begin{align*}
\# {\cal S}(Q,Q) \cdot ( \# \Gamma_{P,R}( \xi ) )^k
& = \# ( {\cal S}(Q, Q) \times \Gamma_{P,R}( \xi )^k ) \\
& = \# M_k \; = \; \# \rho_k( M_k ) \\
& \leq \# ( {\cal S}(Q, Q \oplus S) \times \Gamma_{P,S}( \xi )^k ) \\
& = \# {\cal S}(Q,Q \oplus S) \cdot ( \# \Gamma_{P,S}( \xi ) )^k.
\end{align*}
In the case $\# \Gamma_{P,R}( \xi ) = 1$, \eqref{ungl_GRxi_GSxi} is now trivial, and in the case $\# \Gamma_{P,R}( \xi ) > 1$, \eqref{ungl_GRxi_GSxi} holds because $k \in \mathbb{N}$ is arbitrary.

\end{proof}

The rest of the section is dedicated to the proof of a cancellation rule for I-schemes. In what follows, $Q$, $R$, and $S$ are fixed finite non-empty posets with $Q \oplus R \sqsubseteq_I Q \oplus S$, and the one-to-one homomorphism $\epsilon : {\cal E}( Q \oplus R ) \rightarrow {\cal E}( Q \oplus  S )$ has the property described in Theorem \ref{theo_strongI_equiv_eps}. $X$ is the carrier of $R$ and $Y$ is the carrier of $Q$. Furthermore:
\begin{align*}
{\cal E}(Q)^* & \equiv \mysetdescr{ (x, D, U \cup X) }{(x,D,U) \in {\cal E}(Q) }, \\
{\cal E}(R)^* & \equiv \mysetdescr{ (x, D \cup Y, U ) }{(x,D,U) \in {\cal E}(R) }, \\
{\cal E}(S)^* & \equiv \mysetdescr{ (x, D \cup Y, U ) }{(x,D,U) \in {\cal E}(S) }.
\end{align*}
According to \eqref{EVsys_ordsum}, ${\cal E}(Q)^*, {\cal E}(R)^* \subseteq {\cal E}( Q \oplus R )$ with $\mf{a} <_+ \mf{b} $ for every $\mf{a} \in {\cal E}(Q)^*, \mf{b} \in {\cal E}(R)^*$.

\begin{lemma} \label{lemma_epsER_epsEQ}
We have
\begin{equation} \label{eq_epsER}
\epsilon( {\cal E}(R)^* )_1 \; \subseteq \; S,
\end{equation}
and in the case $Y \subseteq \epsilon( {\cal E}(Q)^* )_1$ we have
\begin{equation} \label{eq_epsERES}
\epsilon( {\cal E}(R)^* ) \; \subseteq \; {\cal E}(S)^*.
\end{equation}
\end{lemma}
\begin{proof} Obviously, $h_{{\cal E}(Q)^*} = h_{{\cal E}(Q)} = h_Q$. Let $\mf{a}^1 <_+ \ldots <_+ \mf{a}^{h_Q}$ be a chain of maximal length in ${\cal E}(Q)^*$ and let $\mf{b} \in {\cal E}(R)^*$, hence $\mf{a}^{h_Q} <_+ \mf{b}$. Then $\epsilon(\mf{a}^1) <_+ \ldots <_+ \epsilon(\mf{a}^{h_Q}) <_+ \epsilon(\mf{b})$ is a chain in ${\cal E}(Q \oplus S)$, and $\epsilon(\mf{a}^1)_1 < \ldots < \epsilon(\mf{a}^{h_Q})_1 < \epsilon(\mf{b})_1$ is a chain in $Q \oplus S$. We conclude $\epsilon(\mf{b})_1 \in S$.

Now assume $Y \subseteq \epsilon( {\cal E}(Q)^* )_1$, and let $\mf{b} \in {\cal E}(R)^*$ be fixed. According to \eqref{eq_epsER}, $\epsilon( \mf{b} )_1 \in S$. Furthermore, due to $\mf{a} <_+ \mf{b}$ for all $\mf{a} \in {\cal E}(Q)^*$, we have $\epsilon( \mf{a} )_1 \in \epsilon( \mf{b} )_2$ for every $\mf{a} \in {\cal E}(Q)^*$. According to our assumption, this means $Y \subseteq \epsilon( \mf{b} )_2$, hence $\epsilon( \mf{b} ) \in {\cal E}( S )^*$.

\end{proof}

We now define two mappings:
\begin{align*}
r : {\cal E}(R) \quad & \rightarrow {\cal E}(Q \oplus R), \\
( x, D, U ) & \mapsto ( x, D \cup Y, U ); \\
s : {\cal E}( Q \oplus S ) & \rightarrow {\cal E}(S), \\
( x, D, U ) & \mapsto ( x, D \setminus Y, U ).
\end{align*}

\begin{corollary} \label{corr_sepsr_is_hom}
Assume $Y \subseteq \epsilon( {\cal E}(Q)^* )_1$. Then
\begin{equation*}
s \circ \epsilon \circ r : {\cal E}(R) \rightarrow {\cal E}(S)
\end{equation*}
is a one-to-one homomorphism.
\end{corollary}
\begin{proof} Obviously, $r$ is an embedding with $r( {\cal E}(R) ) = {\cal E}(R)^*$. For every $\mf{a} \in {\cal E}(R)$, equation \eqref{eq_epsERES} yields $ \epsilon( r( \mf{a} ) ) \in {\cal E}(S)^* $, and because $s$ is one-to-one on ${\cal E}(S)^*$, the total mapping $s \circ \epsilon \circ r$ is one-to-one.

Let $\mf{a}, \mf{a}' \in {\cal E}(R)$ with $ \mf{a} <_+ \mf{a}'$. Because $r$ is an embedding and $\epsilon$ a strict homomorphism, we have $\epsilon(r(\mf{a})) <_+ \epsilon(r(\mf{a}'))$, hence $\epsilon(r(\mf{a}))_1 \in \epsilon(r(\mf{a}'))_2$ and $\epsilon(r(\mf{a}'))_1 \in \epsilon(r(\mf{a}))_3$. Due to $\epsilon(r(\mf{a})), \epsilon(r(\mf{a}')) \in {\cal E}(S)^*$, there exist $ (x , D, U )$, $(x', D', U' ) \in {\cal E}(S)$ with $( x, D \cup Y, U ) = \epsilon( r( \mf{a} ) )$ and $( x', D' \cup Y, U' ) = \epsilon( r( \mf{a}' ) )$. We conclude $ x \in D' $ and $ x' \in U $, thus $s(\epsilon(r( \mf{a} ))) = ( x, D, U ) <_+ ( x', D', U' ) = s(\epsilon(r( \mf{a}' )))$. Hence, $s \circ \epsilon \circ r$ is a homomorphims.

\end{proof}

\begin{lemma} \label{lemma_epsr_23}
Assume $Y \subseteq \epsilon( {\cal E}(Q)^* )_1$. Then
\begin{align} \label{eq_epsraxix_2}
\epsilon( r ( \myaxix{\xi}{x} ) )_2 & =
\mysetdescr{ \epsilon( r ( \myaxix{\xi}{y} ) )_1 }{ y \in {\myodownarrow} \gxix } \; \cup \; Y ,\\ \label{eq_epsraxix_3}
\epsilon( r ( \myaxix{\xi}{x} ) )_3 & =
\mysetdescr{ \epsilon( r ( \myaxix{\xi}{y} ) )_1 }{ y \in {\myouparrow} \gxix },
\end{align}
for all $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$. In \eqref{eq_epsraxix_2}, the two sets on the right are disjoint.
\end{lemma}
\begin{proof} Let $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$ be fixed. Due to $r ( \myaxix{\xi}{x} ) = \myapxix{Q \oplus P}{ id_Q \cup \xi}{x}$ we get with $\eta$ defined as in Theorem \ref{theo_strongI_equiv_eps}
\begin{align*}
\epsilon( r ( \myaxix{\xi}{x} ) ) & = \epsilon( \myapxix{Q \oplus P}{ id_Q \cup \xi}{x} ) \; = \; \myapxix{Q \oplus P}{ \eta( id_Q \cup \xi )}{x}.
\end{align*}
The mapping $\eta$ is an I-scheme and thus a G-scheme. Therefore, $\mygxix{\eta(\zeta)}{z} = \mygxix{\zeta}{z}$ for all $\zeta \in {\cal H}(Q \oplus P, Q \oplus R)$, $z \in Q \oplus P$, and thus
\begin{align*}
\epsilon( r ( \myaxix{\xi}{x} ) )_2 & = \myapxix{Q \oplus P}{ \eta( id_Q \cup \xi )}{x}_2 \\
& = \eta( id_Q \cup \xi )\left( \mydngxix{id_Q \cup \xi}{x} \right) \\
& = \mysetdescr{ \eta( id_Q \cup \xi )(y) }{ y \in \mydngxix{id_Q \cup \xi}{x} } \\
& = \mysetdescr{ \epsilon( \myapxix{Q \oplus P}{ id_Q \cup \xi}{y} )_1 }{ y \in \mydngxix{id_Q \cup \xi}{x} }.
\end{align*}
Because of $x \in R$, we have ${\downarrow^{^{\!\!\!\circ}}_{Q \oplus P}} \mygxix{id_Q \cup \xi}{x} = Y \cup {\downarrow^{^{\!\!\!\circ}}_{P}} \mygxix{\xi}{x}$. With
\begin{align*}
N_Q & \equiv \mysetdescr{ \epsilon( \myapxix{Q \oplus P}{ id_Q \cup \xi}{y} )_1 }{ y \in Y }, \\
N_P & \equiv \mysetdescr{ \epsilon( \myapxix{Q \oplus P}{ id_Q \cup \xi}{y} )_1 }{ y \in {\downarrow^{^{\!\!\!\circ}}_{P}} \mygxix{\xi}{x} }
\end{align*}
we have thus $\epsilon( r( \myaxix{\xi}{x} ) )_2 = N_Q \cup N_P$.

Let $y \in {\downarrow^{^{\!\!\!\circ}}_{P}} \mygxix{\xi}{x}$, hence $y \in P$. Then we have $Y \subseteq  \myapxix{Q \oplus P}{ id_Q \cup \xi}{y}_2$, thus $ \myapxix{Q \oplus P}{ id_Q \cup \xi}{y} \in {\cal E}(R)^*$. Equation \eqref{eq_epsER} delivers $\epsilon( \myapxix{Q \oplus P}{ id_Q \cup \xi}{y})_1 \in S$, and we conclude $N_P \subseteq S$. In particular, $N_P \cap Y = \emptyset$.

Due to \eqref{eq_epsERES}, we have $ Y \subseteq \epsilon( r( \myaxix{\xi}{x} ) ) )_2$. We conclude $Y \subseteq N_Q$, and with $ \# N_Q \leq \# Y$ we get $Y = N_Q$. Now \eqref{eq_epsraxix_2} is shown.

In the same way we see
\begin{align*}
\epsilon( r ( \myaxix{\xi}{x} ) )_3 & = \mysetdescr{ \epsilon( \myapxix{Q \oplus P}{ id_Q \cup \xi}{y} )_1 }{ y \in \myupgxix{id_Q \cup \xi}{x} }
\end{align*}
and \eqref{eq_epsraxix_3} follows because of ${{\uparrow_{_{_{\!\!\!\circ}}}}_{Q \oplus P}} \mygxix{id_Q \cup \xi}{x} = {{\uparrow_{_{_{\!\!\!\circ}}}}_{P}} \mygxix{\xi}{x}$.

\end{proof}

\begin{theorem} \label{theo_cancel_ordsumI}
Assume $Y \subseteq \epsilon( {\cal E}(Q)^* )_1$. Then $R \sqsubseteq_I S$. In particular, the mapping
\begin{align*}
T \equiv s \circ \epsilon \circ r : {\cal E}(R) \rightarrow {\cal E}(S)
\end{align*}
is a one-to-one homomorphism which fulfills the requirement of Theorem \ref{theo_strongI_equiv_eps}: for all $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$
\begin{align*}
\myaxix{\tau_P(\xi)}{x} & = T( \myapxix{P}{\xi}{x} ), \\
\mytext{where} \quad
\tau_P(\xi)(x) & \equiv T( \myapxix{P}{\xi}{x} )_1.
\end{align*}
\end{theorem}
\begin{proof} In Corollary \ref{corr_sepsr_is_hom}, we have seen that $T$ is a one-to-one homomorphism.
$\myaxix{\tau(\xi)}{x}_1 = T( \myaxix{\xi}{x} )_1$ is trivial. Due to Lemma \ref{lemma_exix},  $\mygxix{\tau(\xi)}{x} = \mygxix{\xi}{x}$, and with $\tau(\xi)(x) = T( \myaxix{\xi}{x} )_1 = \epsilon( r( \myaxix{\xi}{x} ) )_1$ we get
\begin{align*}
\myaxix{\tau(\xi)}{x}_2 & = \tau(\xi)( {\myodownarrow} \gxix ) = \mysetdescr{ \epsilon( r ( \myaxix{\xi}{y} ) )_1 }{ y \in {\myodownarrow} \gxix } \\
& \stackrel{\eqref{eq_epsraxix_2}}{=} \epsilon( r ( \myaxix{\xi}{x} ) )_2 \setminus Y = s( \epsilon( r( \myaxix{\xi}{x} ) ) )_2 = T( \myaxix{\xi}{x} )_2.
\end{align*}
$ \myaxix{\tau(\xi)}{x}_3 = T( \myaxix{\xi}{x} )_3$ is shown in the same way by using \eqref{eq_epsraxix_3}.

\end{proof}

The dual results are

\begin{theorem} \label{theo_cancel_ordsum_duals}
Let $Q, R, S \in  \myfr{P} $ with $Q \cap R = \emptyset$ and $Q \cap S = \emptyset$. Then
\begin{equation*}
R \oplus Q \sqsubseteq_G S \oplus Q \quad \Rightarrow \quad R \sqsubseteq_G S.
\end{equation*}
In the case $R \oplus Q \sqsubseteq_I S \oplus Q$, let $\epsilon : {\cal E}( R \oplus Q ) \rightarrow {\cal E}( S \oplus Q )$ be the one-to-one homomorphism from Theorem \ref{theo_strongI_equiv_eps}, and let
\begin{equation*}
{\cal E}(Q)_* \equiv \mysetdescr{ (x, D \cup X, U) }{(x,D,U) \in {\cal E}(Q) },
\end{equation*}
where $X$ is the carrier of $R$. With $Y$ being the carrier of $Q$, we have
\begin{equation*}
Y \subseteq \epsilon( {\cal E}(Q)_* )_1 \quad \Rightarrow \quad R \sqsubseteq_I S.
\end{equation*}
\end{theorem}

\section{Cancellation rules for products} \label{sec_cancel_product}

In this section, $Q, R, S \in  \myfr{P} $ are fixed posets, and $\rho$ is a strong Hom-scheme from $Q \times R$ to $Q \times S$. The first cancellation rule is easily proven: For all $P \in  \myfr{P} _r$ we have
\begin{align*}
\# {\cal H}(P,Q) \cdot \# {\cal H}(P,R) & = \# {\cal H}(P, Q \times R) \; \leq \; \# {\cal H}(P, Q \times S) \\
& = \# {\cal H}(P,Q) \cdot \# {\cal H}(P,S),
\end{align*}
thus $R \sqsubseteq S$.

In the proofs of the cancellation rules for strong G-schemes and I-schemes, constant homomorphisms will play an important role:
\begin{definition} \label{def_CP_cPq}
For all $P \in  \myfr{P} $ and all $q \in Q$, let $c_{P,q} \in {\cal H}( P, Q )$ be the homomorphism mapping all $x \in P$ to $q$.
\begin{equation*}
{\cal C}(P) \; \equiv \; \mysetdescr{ c_{P,q} \in {\cal H}(P,Q) }{ q \in Q }
\end{equation*}
is the set of the constant homomorphims from $P$ to $Q$.
\end{definition}

\begin{corollary} \label{corr_cqxi}
Let $P \in  \myfr{P} $, $q \in Q$, and $\xi \in {\cal H}(P,R)$. Then for all $x \in P$
\begin{align} \label{eq_gxix_cqxi}
\mygxix{(c_{P,q},\xi)}{x} & = \mygxix{\xi}{x}, \\
\myaxix{(c_{P,q},\xi)}{x} & = \label{eq_axix_cqxi}
\big( ( q, \xi(x) ),
\{ q \} \times \myaxix{\xi}{x}_2,
\{ q \} \times \myaxix{\xi}{x}_3 \big).
\end{align}
\end{corollary}
\begin{proof} $\mygxix{c_{P,q}}{x}$ is for every $x \in P$ the connectivity component of $P$ containing $x$, thus $\mygxix{\xi}{x} \subseteq \mygxix{c_{P,q}}{x}$ because $\mygxix{\xi}{x}$ is connected with $x \in \mygxix{\xi}{x}$. Hence,
\begin{align*}
\mygxix{ ( c_{P,q}, \xi) }{x} & \stackrel{\eqref{gxix_produkt}}{=}
\gamma_{\mygxix{c_{P,q}}{x} \cap \mygxix{\xi}{x}}(x)
= \gamma_{\mygxix{\xi}{x}}(x) \stackrel{\eqref{gxix_gammagxix}}{=} \mygxix{\xi}{x}.
\end{align*}
$ \myaxix{(c_{P,q},\xi)}{x}_1 = (q, \xi(x) )$ is trivial. Furthermore,
\begin{align*}
\myaxix{(c_{P,q},\xi)}{x}_2 & =
( c_{P,q}, \xi)\Big( {\downarrow^{^{\!\!\!\circ}}} \mygxix{ ( c_{P,q}, \xi) }{x} \Big)
\stackrel{\eqref{eq_gxix_cqxi}}{=}
( c_{P,q}, \xi)\Big( {\downarrow^{^{\!\!\!\circ}}} \mygxix{\xi}{x} \Big) \\
& =
\{ q \} \times \xi\Big( {\downarrow^{^{\!\!\!\circ}}} \mygxix{\xi}{x} \Big) =
\{ q \} \times \myaxix{\xi}{x}_2
\end{align*}
and similarly $\myaxix{(c_{P,q},\xi)}{x}_3 = \{ q \} \times \myaxix{\xi}{x}_3$.

\end{proof}

In this section, the key is the following set-theoretical lemma:

\begin{lemma} \label{lemma_produktzyklus}
Let $A, B, C$ be non-empty sets, $A$ finite, and let $f : A \times B \rightarrow C$ and $g : C \rightarrow A$ be mappings with
\begin{equation} \label{bed_komp_1_allg}
\forall \; a, a' \in A, b \in B \; : \quad a \not= a' \; \Rightarrow \; g(f(a,b)) \not= g(f(a',b)).
\end{equation}
For a fixed $a_0 \in A$, we define recursively for every $b \in B$ the sequence
\begin{align*}
\chi^0 (b) & \equiv a_0, \\
\forall \; i \in \mathbb{N} \; : \; \chi^i (b) & \equiv g( f ( \chi^{i-1}(b), b)).
\end{align*}
Then, for every $b \in B$, there exists an integer $n(b) \in \mathbb{N}$ with
\begin{align*}
\chi^{n(b)} (b) & = a_0.
\end{align*}
\end{lemma}
\begin{proof} Let $b \in B$. We have $\chi^i(b) \in A$ for every $i \in \mathbb{N}_0$, and because $A$ is finite, there exist $i, j \in \mathbb{N}_0$ with $i > j$ and $\chi^i(b) = \chi^j(b)$. In the case $j = 0$, we have $\chi^i(b) = a_0$. For $j > 0$ we get
\begin{align*}
g(f( \chi^{i-1}(b), b)) & = \; \chi^i(b) \spc{=} \chi^j(b) \spc{=} g(f( \chi^{j-1}(b), b )),
\end{align*}
and with \eqref{bed_komp_1_allg} we conclude $\chi^{i-1}(b) = \chi^{j-1}(b)$. Stepping backwards in this way, we finally reach $\chi^{i-j}(b) = \chi^0(b) = a_0$.

\end{proof}

Now let $\rho$ be a G-scheme from $Q \times R$ to $Q \times S$. In order to be able to show $R \sqsubseteq_G S$, we will assume that $\rho$ fulfills \eqref{bed_komp_1_allg}:
\begin{align} \label{bed_komp_1}
\forall P \in  \myfr{P} ,  \xi, \zeta \in {\cal H}(P,Q), \theta \in {\cal H}(P,R) & :
\\
\xi \not= \zeta \; & \Rightarrow \; \rho( \xi, \theta )_1 \not= \rho( \zeta, \theta )_1, \nonumber
\end{align}
thus, (with the symbols used in Lemma \ref{lemma_produktzyklus}): $A = {\cal H}(P,Q)$, $B = {\cal H}(P,R)$, $C = {\cal H}(P, Q \times S)$, $f = \rho_P$, $g = \pi_1$. The counterpart to the sequences $\chi^i(b)$ is defined as follows:
\begin{definition} \label{def_phi}
Let $q \in Q$ be fixed. For every $\xi \in {\cal H}(P, R )$, we define recursively
\begin{align*}
\phi_P^0(\xi) & \equiv c_{P,q} \\
\forall \; i \in \mathbb{N} \; : \;
\phi_P^{i}(\xi) & \equiv \rho( \phi_P^{i-1}(\xi), \xi )_1.
\end{align*}
As complementary notation, we define
\begin{align*}
\forall \; i \in \mathbb{N}_0 \; : \;
\Phi_P^{i}(\xi) & \equiv ( \phi_P^i(\xi), \xi ) =
\begin{cases}
( c_{P,q}, \xi ), & \mytext{if} i = 0; \\
\left( \rho( \Phi_P^{i-1}(\xi) )_1, \xi \right), & \mytext{if} i \geq 1.
\end{cases}
\end{align*}
\end{definition}

\begin{lemma} \label{lemma1_phi_zu_QxR}
For every $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$, and every $i \in \mathbb{N}_0$
\begin{align} \label{grhophix_eq_gxix_1}
\mygxix{ \Phi_P^i(\xi) }{x} & = \mygxix{\xi}{x}.
\end{align}
\end{lemma}
\begin{proof} Let $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$ be selected. In order to unburden the notation, we skip the index $P$. $\mygxix{ \Phi^0(\xi) }{x} = \mygxix{\xi}{x}$ holds due to Corollary \ref{corr_cqxi}.

Assume that \eqref{grhophix_eq_gxix_1} holds for $i \in \mathbb{N}_0$. Then $\mygxix{\xi}{x} = \mygxix{ \Phi^i(\xi) }{x}$ $= \mygxix{ \rho( \Phi^i(\xi)) }{x}$ $\subseteq \mygxix{\rho( \Phi^{i}(\xi) )_1}{x}$ according to \eqref{gxix_produkt}, thus
\begin{align*}
\mygxix{ \Phi^{i+1}(\xi) }{x} & =
\mygxix{ ( \rho( \Phi^{i}(\xi) )_1, \xi) }{x}
\stackrel{\eqref{gxix_produkt}}{=}
\gamma_{\mygxix{\rho( \Phi^{i}(\xi))_1}{x} \; \cap \; \mygxix{\xi}{x}}(x) \\
& = \gamma_{\mygxix{\xi}{x}}(x) \stackrel{\eqref{gxix_gammagxix}}{=} \mygxix{\xi}{x}.
\end{align*}

\end{proof}

\begin{theorem} \label{theo_cancel_prod_G}
Let $\rho$ fulfill \eqref{bed_komp_1}. According to Lemma \ref{lemma_produktzyklus}, there exists for every $\xi \in {\cal H}(P,R)$ an integer $n_P(\xi) \in \mathbb{N}$ with
\begin{align*}
\phi_P^{n_P(\xi)}(\xi) & = c_{P,q}.
\end{align*}
Then the mapping
\begin{align*}
\tau & \in \prod_{P \in  \myfr{P} _r} {\cal A}( {\cal H}(P,R), {\cal H}(P,S) ) \\
\mytext{defined by} \quad
\tau_P(\xi) & \equiv \rho( \Phi_P^{n_P(\xi)-1}(\xi) )_2
\end{align*}
for all $P \in  \myfr{P} _r, \xi \in {\cal H}(P,R)$, is a strong G-scheme from $R$ to $S$.
\end{theorem}
\begin{proof} Let $P \in \mfP_r$, $\xi \in {\cal H}(P,R)$, $x \in P$. Skipping the index $P$, we have
$\rho( \Phi^{n(\xi)-1}(\xi))_1 = \phi^{n(\xi)}(\xi) = c_{P,q}$, thus
\begin{align*}
\mygxix{ \tau( \xi ) }{x} &
\stackrel{\eqref{eq_gxix_cqxi}}{=}
\mygxix{ \rho( \Phi^{n(\xi)-1}(\xi)) }{x} =
\mygxix{ \Phi^{n(\xi)-1}(\xi) }{x}
\stackrel{\eqref{grhophix_eq_gxix_1}}{=} \mygxix{\xi}{x},
\end{align*}
and $\tau$ is a G-scheme.

Let $\xi, \zeta \in {\cal H}(P,R)$ with $\tau(\xi) = \tau(\zeta)$. Then
\begin{align*}
\rho( \Phi^{n(\xi)-1}(\xi)_1, \xi ) & =
( c_{P,q}, \tau(\xi) )
= ( c_{P,q}, \tau(\zeta) )
= \rho( \Phi^{n(\zeta)-1}(\zeta)_1, \zeta ),
\end{align*}
and because $\rho$ is one-to-one we get $ ( \Phi^{n(\xi)-1}(\xi)_1, \xi ) = ( \Phi^{n(\zeta)-1}(\zeta)_1, \zeta )$, thus $\xi = \zeta$, and $\tau_P$ is one-to-one.

\end{proof}

Also in this section, the proof of the cancellation rule for I-schemes will be the more complicated one. Assume $ Q \times R \sqsubseteq_I Q \times S $, and let $\rho$ be a strong I-scheme from $R$ to $S$. For the proof of $R \sqsubseteq_I S$, we need an additional assumption about $\rho$:
\begin{align} \label{assumption_Ischeme}
\forall \; \xi \in {\cal H}(P, Q \times R) & \; : \;
\xi_1 \in {\cal C}(P) \; \Rightarrow \; \rho_P(\xi)_1 \in {\cal C}(P).
\end{align}

\begin{lemma} \label{lemma_alpha_Phi}
Let $\rho$ fulfill \eqref{assumption_Ischeme}. Then for all $P, P' \in  \myfr{P} _r, \xi \in {\cal H}(P, Q \times R)$, $\zeta \in {\cal H}(P', Q \times R)$, $x \in P$, $y \in P'$
\begin{align} \label{aPhiixi_le_aPhiizeta}
\myapxix{P}{\xi}{x} \leq_+ \myapxix{P'}{\zeta}{y} & \; \Rightarrow \;
\forall \; i \in \mathbb{N}_0 \; : \; \myapxix{P}{\Phi_P^i(\xi)}{x} \leq_+ \myapxix{P'}{\Phi_{P'}^i(\zeta)}{y},
\end{align}
where ``$<_+ / = $'' on the left side implies ``$<_+ / =$'' on the right side. Additionally,
\begin{align} \label{Phiixi_eq_Phiizeta}
\myapxix{P}{\xi}{x} \leq_+ \myapxix{P'}{\zeta}{y} & \; \Rightarrow \;
\forall \; i \in \mathbb{N}_0 \; : \; \Phi_P^i(\xi)_1 = \Phi_{P'}^i(\zeta)_1
\end{align}
for every $P, P' \in  \myfr{P} _r$, $\xi \in {\cal H}(P, Q \times R)$, $\zeta \in {\cal H}(P', Q \times R)$, $x \in P$, and $y \in P'$.
\end{lemma}
\begin{proof} Also in this proof, we skip the indices $P$ and $P'$. For $\myaxix{\xi}{x} = \myaxix{\zeta}{y}$, \eqref{eq_axix_cqxi} delivers $\myaxix{\Phi^0(\xi)}{x} = \myaxix{\Phi^0(\zeta)}{y}$. Assume $\myaxix{\xi}{x} <_+ \myaxix{\zeta}{y}$. Then $\xi(x) \in \myaxix{\zeta}{y}_2$ and $\zeta(y) \in \myaxix{\xi}{x}_3$, thus
\begin{align*}
\Phi^0(\xi)(x) & = ( q, \xi(x) ) \in \{ q \} \times \myaxix{\zeta}{y}_2
\stackrel{\eqref{eq_axix_cqxi}}{=} \myaxix{\Phi^0(\zeta)}{y}_2
\end{align*}
and similarly $ \Phi^0(\zeta)(y) \in \myaxix{\Phi_P^0(\xi)}{x}_3$. Therefore, $\myaxix{\Phi^0(\xi)}{x} <_+ \myaxix{\Phi^0(\zeta)}{y}$, and \eqref{aPhiixi_le_aPhiizeta} holds for $i = 0$. \eqref{Phiixi_eq_Phiizeta} is trivial for $i = 0$.

Assume that $\myaxix{\Phi^i(\xi)}{x} \leq_+ \myaxix{\Phi^i(\zeta)}{y}$ holds for $ i \in \mathbb{N}_0 $. Because $\rho$ is an I-scheme, we have $\myarxix{\Phi^i(\xi)}{x}$ $\leq_+$ $\myarxix{\Phi^i(\zeta)}{y}$. By applying \eqref{assumption_Ischeme} $(i+1)$ times we see that $ \rho(\Phi^i(\xi))_1 $, and $ \rho(\Phi^i(\zeta))_1 $ are both elements of ${\cal C}(P)$. There exist thus $r, s \in Q$ with $ \rho(\Phi^i(\xi))_1 = c_{P,r} $ and $ \rho(\Phi^i(\zeta))_1 = c_{P',s} $.

In the case $\myaxix{\xi}{x} = \myaxix{\zeta}{y}$, we have $\myaxix{\Phi^i(\xi)}{x} = \myaxix{\Phi^i(\zeta)}{y}$, and \eqref{eq_imagebased_rhowert} yields $\myarxix{\Phi^i(\xi)}{x}$ $=$ $\myarxix{\Phi^i(\zeta)}{y}$. Now
\begin{align*}
( r, \rho(\Phi^i(\xi))_2(x) ) & = \rho(\Phi^i(\xi))(x) = \rho(\Phi^i(\zeta))(y) = ( s, \rho(\Phi^i(\zeta))_2(y) ),
\end{align*}
thus $r = s$. And in the case $\myaxix{\xi}{x} <_+ \myaxix{\zeta}{y}$, we have $\myaxix{\Phi^i(\xi)}{x} <_+ \myaxix{\Phi^i(\zeta)}{y}$, hence $\myarxix{\Phi^i(\xi)}{x}$ $<_+$ $\myarxix{\Phi^i(\zeta)}{y}$. Therefore,
\begin{align*}
( r, \rho(\Phi^i(\xi))_2(x) ) & =  \rho(\Phi^i(\xi))(x) \in
\myarxix{\Phi^i(\zeta)}{y}_2 \stackrel{\eqref{eq_axix_cqxi}}{=}
 \{ s \} \times \myaxix{\rho(\Phi^i(\zeta))_2}{y}_2,
\end{align*}
hence $r = s$ also in this case. \eqref{Phiixi_eq_Phiizeta} holds thus for $i+1$, and we get with $\xi(x) \in \myaxix{\zeta}{y}_2$
\begin{align*}
\Phi^{i+1}(\xi)(x) & = ( r, \xi(x) ) \in \{ r \} \times \zeta \left( \downarrow^{^{\!\!\!\circ}} \mygxix{\zeta}{y} \right) =
\{ s \} \times \zeta \left( \downarrow^{^{\!\!\!\circ}} \mygxix{\zeta}{y} \right) \\
& = ( c_{P,s}, \zeta ) \left( \downarrow^{^{\!\!\!\circ}} \mygxix{\zeta}{y} \right)
\stackrel{\eqref{grhophix_eq_gxix_1}}{=}
\myaxix{\Phi^{i+1}(\zeta)}{y}_2,
\end{align*}
and in the same way we prove $ \Phi^{i+1}(\zeta)(y) \in \myaxix{\Phi^{i+1}(\xi)}{x}_3$. Thus, $\myaxix{\Phi^{i+1}(\xi)}{x}$ $<_+$ $\myaxix{\Phi^{i+1}(\zeta)}{y}$.

\end{proof}

\eqref{Phiixi_eq_Phiizeta} shows that assumption \eqref{assumption_Ischeme} is restrictive. For all $i \in \mathbb{N}_0$, it enforces $\phi_P^i(\xi) = \phi_P^i(\zeta)$ for all $\xi, \zeta \in {\cal H}(P, Q \times R)$ for which $\myapxix{P}{\xi}{x}$ and $\myapxix{P}{\zeta}{y}$ belong to the same connectivity component of ${\cal E}(Q \times R)$ for some $x, y \in P$.

\begin{theorem} \label{theo_cancel_prod_I}
Let $\rho$ fulfill \eqref{bed_komp_1} and \eqref{assumption_Ischeme}.
Then $R \sqsubseteq_I S$, and the mapping $\tau$ defined in Theorem \ref{theo_cancel_prod_G} is a strong I-scheme from $R$ to $S$.
\end{theorem}
\begin{proof} Let $P \in  \myfr{P} _r, \xi \in {\cal H}(P,Q \times R)$, and let $n_P(\xi)$ be defined as in Theorem \ref{theo_cancel_prod_G}. Due to $\phi^{n_P(\xi)}( \xi ) = c_{P,q}$, we have $ \Phi^{n_P(\xi)}( \xi ) = ( c_{P,q}, \xi ) = \Phi^0( \xi )$. We conclude $ \Phi^{k \cdot n_P(\xi) - 1}( \xi ) = \Phi^{n_P(\xi) - 1}( \xi )$ for all $k \in \mathbb{N}$.

Let $P, P' \in  \myfr{P} _r, \xi \in {\cal H}(P, Q \times R)$, $\zeta \in {\cal H}(P', Q \times R)$, $x \in P$, $y \in P'$ with $\myaxix{\xi}{x} \leq_+ \myaxix{\zeta}{y}$. (Now we skip again the indices $P$ and $P'$.) With $N \equiv \lcm \{n(\xi), n(\zeta) \}$, Lemma \ref{lemma_alpha_Phi} yields $\myaxix{\Phi^{N-1}(\xi)}{x} \leq_+ \myaxix{\Phi^{N-1}(\zeta)}{y}$ with ``$<_+$'' if $\myaxix{\xi}{x} <_+ \myaxix{\zeta}{y}$ and with ``$=$'' if $\myaxix{\xi}{x} = \myaxix{\zeta}{y}$. Because $\rho$ is an I-scheme, we get
\begin{align*}
\myarxix{\Phi^{n(\xi)-1}(\xi)}{x}
& =  \myarxix{\Phi^{N-1}(\xi)}{x}
\leq_+ \myarxix{\Phi^{N-1}(\zeta)}{y}
= \myarxix{\Phi^{n(\zeta)-1}(\zeta)}{y},
\end{align*}
again with ``$<_+$'' if $\myaxix{\xi}{x} <_+ \myaxix{\zeta}{y}$, and with ``$=$'' if $\myaxix{\xi}{x} = \myaxix{\zeta}{y}$, according to \eqref{eq_imagebased_rhowert}.

Assume $\myaxix{\xi}{x} = \myaxix{\zeta}{y}$. Then, with Corollary \ref{corr_cqxi}
\begin{align*}
& \; \big( ( q, \myaxix{\tau(\xi)}{x}_1 ),
\{ q \} \times \myaxix{\tau(\xi)}{x}_2,
\{ q \} \times \myaxix{\tau(\xi)}{x}_3 \big) \\
= & \; \myarxix{\Phi^{n(\xi)-1}(\xi)}{x}
= \myarxix{\Phi^{n(\xi)-1}(\zeta)}{y} \\
= &  \; \big( ( q, \myaxix{\tau(\zeta)}{y}_1 ),
\{ q \} \times \myaxix{\tau(\zeta)}{y}_2,
\{ q \} \times \myaxix{\tau(\zeta)}{y}_3 \big),
\end{align*}
thus $\myaxix{\tau(\xi)}{x} = \myaxix{\tau(\zeta)}{y}$. And in the case $\myaxix{\xi}{x} <_+ \myaxix{\zeta}{y}$, we get again with Corollary \ref{corr_cqxi}
\begin{align*}
( q, \tau(\xi)(x) ) &
= \myarxix{\Phi^{n(\xi)-1}(\xi)}{x}_1
\in \myarxix{\Phi^{n(\zeta)-1}(\zeta)}{y}_2
= \{ q \} \times \myaxix{\tau(\zeta)}{y}_2,
\end{align*}
thus $\tau(\xi)(x) \in \myaxix{\tau(\zeta)}{y}_2$. In the same way we see $\tau(\zeta)(y) \in \myaxix{\tau(\xi)}{x}_3$, and $\myaxix{\tau(\xi)}{x} <_+ \myaxix{\tau(\zeta)}{y}$ is proven. $\tau$ is thus an I-scheme, and it is strong according to Theorem \ref{theo_cancel_prod_G}.

\end{proof}

redpajama_set_name: RedPajamaArXiv