chap-tensor.tex

\chapter{Matters related to tensor product}

These consideration on (possibly infinite) indexed families of
join-semilattices is based on \cite{nforum-todd-tensor} (for the finite case).

Let $\mathfrak{A}$ be an indexed family of join-semilattices with least
elements. Let $T$ also be a join-semilattice.

Let $F (X)$ mean free join-semilattice for a set $X$.

\begin{defn}
$\mathbf{SepJoin}(\prod\mathfrak{A},T)$ is the set of maps from $\prod\mathfrak{A}$
to~$T$, preserving joins in every argument $i\in\dom\mathfrak{A}$.
\end{defn}

\begin{obvious}
The set of free join-semilattices $F (X)$ is order-isomorphic to the set of
subsets $X$ of a ``universal'' set $\mho$.{\hspace*{\fill}}{\medskip}
\end{obvious}

Let $i : \prod \mathfrak{A} \rightarrow F \left( \prod \mathfrak{A} \right)$
be the universal embedding.

Let $\sim$ be defined as the smallest equivalence relation on $F \left( \prod
\mathfrak{A} \right)$ that for every $k \in \dom \mathfrak{A}$, $L \in
\prod_{i \in (\dom \mathfrak{A}) \setminus \{ k \}} \mathfrak{A}_i$:
\begin{enumerate}
  \item $i (L \cup \{ (k , g \sqcup h) \}) \sim i (L \cup \{ (k , g) \})
  \sqcup i (L \cup \{ (k , h) \})$;
  
  \item $\bot \sim i (L \cup \{ (k , \bot) \})$;
  
  \item $x \sim y \wedge x' \sim y' \Rightarrow x \sqcup x' \sim y \sqcup y'$
  for all $x, y, x', y' \in F \left( \prod \mathfrak{A} \right)$.
\end{enumerate}

\begin{obvious}
Some function $h : X \rightarrow Y$ induces a well defined map $\psi : X / E
\rightarrow Y$ on equivalence classes, if $E \subseteq F$ where $x \mathrel{F}
y \Leftrightarrow h x = h y$.{\hspace*{\fill}}{\medskip}
\end{obvious}

\begin{lem}
  The set of join-homomorphisms $\psi : F \left( \prod \mathfrak{A} \right) / \sim
  \rightarrow T$ is isomorphic to the set of maps $\phi :
  \prod \mathfrak{A} \rightarrow T$ preserving finite joins in separate arguments.
\end{lem}

\begin{proof}
  The quotient map $q : F \left( \prod \mathfrak{A} \right) \rightarrow F
  \left( \prod \mathfrak{A} \right) / \sim$ which takes an element $x$ to its
  equivalence class $[x]$ map is well defined because
  \[ x \sim y \wedge x' \sim y' \Rightarrow x \sqcup x' \sim y \sqcup y' . \]
  The map $q$ \ preserves join. $F \left( \prod \mathfrak{A} \right) / \sim$
  is associative, commutative, and idempotent since it is so on $F \left(
  \prod \mathfrak{A} \right)$ and thus is a join-semilattice.
  
  Let join-preserving map $\psi : F \left( \prod \mathfrak{A} \right) / \sim
  \rightarrow T$. It is easy to show that $\psi \circ q \circ i$ preserves
  joins in separate arguments.
  
  Let now $\phi : \prod \mathfrak{A} \rightarrow T$ preserves joins in
  separate arguments. There is a unique join-preserving map $\tilde{\phi} : F
  \left( \prod \mathfrak{A} \right) \rightarrow T$ such that $\tilde{\phi}
  \circ i = \phi$. We must show that this induces a well-defined
  join-preserving map $\psi : F \left( \prod \mathfrak{A} \right) / \sim
  \rightarrow T$ such that $\psi (q (x)) = \tilde{\phi} (x)$ for all $x \in F
  \left( \prod \mathfrak{A} \right)$ (clearly at most one function $\psi$ can
  satisfy this equation since $q$ is surjective). This will show that $\psi$
  bijectively correspond to $\tilde{\phi}$ and thus bijectively correspond to
  $\phi$. (This will finish the proof as that this bijection is monotone is
  obvious.)
  
  Using the ``obvious'' above, it's enough (taking into account that $\sim$ is
  the minimal equivalence relation subject to the above formulas) to prove
  that:
  \begin{enumerate}
    \item $\tilde{\phi} (i (L \cup \{ (k , g \sqcup h) \})) = \tilde{\phi} (i
    (L \cup \{ (k , g) \}) \sqcup i (L \cup \{ (k , h) \}))$;
    
    \item $\tilde{\phi} (\bot) = \tilde{\phi} (i (L \cup \{ (k , \bot) \}))$;
    
    \item $\tilde{\phi} (x) = \tilde{\phi} (y) \wedge \tilde{\phi} (x') =
    \tilde{\phi} (y') \Rightarrow \tilde{\phi} (x \sqcup x') = \tilde{\phi} (y
    \sqcup y')$
  \end{enumerate}
  The first easily follows from $\tilde{\phi} \circ i = \phi$ and the fact
  that $\tilde{\phi}$ preserves binary joins.
  
  The second easily follows from $\tilde{\phi} \circ i = \phi$ and that $\phi$
  preserves $\bot$.
  
  The third follows from the fact that $\tilde{\phi}$ preserves joins.
\end{proof}

\begin{cor}
  The poset of prestaroids
  $\mathsf{preStrd} (\mathfrak{A})$ is isomorphic to an ideal
  (on a join-semilattice), provided that $\mathfrak{A}$ is an indexed family
  of join-semilattices.
\end{cor}

\begin{proof}
  $\mathsf{preStrd} (\mathfrak{A}) \cong \mathbf{SepJoin} (\mathfrak{A}, 2)
  \cong F \left( \prod \mathfrak{A} \right) / \sim \rightarrow 2 \cong
  \mathfrak{I} \left( F \left( \prod \mathfrak{A} \right) / \sim \right)$.
\end{proof}

\fxwarning{Check below (especially posets vs dual posets) for errors.}

\begin{cor}
$\mathsf{preStrd}$ is a complete lattice.
\end{cor}

\begin{proof}
Corollary~\bookref{filt-is-complete}.
\end{proof}

\begin{cor}
$\mathsf{preStrd}$ is a filtered filtrator.
\end{cor}

\begin{proof}
Theorem~\bookref{semifilt-joinclosed}.
\end{proof}

\fxnote{Try to prove that $\mathsf{preStrd}$ is atomic and moreover atomistic (under certain conditions). Other properties?}