First draft for open preimage functions.

src/topology/main/index.tex

@@ -26,3 +26,4 @@
 \input{./baire.tex}
 \input{./cube.tex}
 \input{./compactify.tex}
+\input{./preimage.tex}

src/topology/main/preimage.tex

@@ -0,0 +1,111 @@
+\section{Open Preimage Functions}
+\label{section:preimage-function-topology}
+
+\begin{definition}[Open Preimage Function]
+\label{definition:open-preimage-function}
+    Let $X$ be a set, $(Y, \topo)$ be a topological space, and $P: \topo \to 2^X$, then $P$ is an \textbf{open preimage function} if
+    \begin{enumerate}[label=(PF\arabic*)]
+        \item $P(\emptyset) = \emptyset$. 
+        \item[(PF2')] For each $\mathcal{U} \subset \topo$, $\bigcup_{U \in \mathcal{U}}P(U) = P\paren{\bigcup_{U \in \mathcal{U}}U}$. 
+        \item[(PF3')] For each $U, V \in \topo$, $P(U \cap V) = P(U) \cap P(V)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:open-preimage-function-gymnastics}
+    Let $X$ be a set, $(Y, \topo)$ be a topological space, and $f: X \to Y$, then:
+    \begin{enumerate}
+        \item The mapping $U \mapsto f^{-1}(U)$ is an open preimage function. 
+        \item If $Y$ is T1, then for any $g: X \to Y$ with $g^{-1}(U) = f^{-1}(U)$ for all $U \in \topo$, $f = g$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Let $y \in Y$, then since $Y$ is T1, $\bracs{x} = \bigcap_{U \in \topo, y \in U}U$, so
+    \[
+    g^{-1}(\bracs{y}) = \bigcap_{\substack{U \in \topo \\ y \in U}}g^{-1}(U) = \bigcap_{\substack{U \in \topo \\ y \in U}}f^{-1}(U) = f^{-1}(\bracs{y})
+    \]
+    
+    Thus for any $x \in X$, $f(x) = y$ if and only if $g(x) = y$, so $f = g$. 
+\end{proof}
+
+
+\begin{definition}[Basic Preimage Function]
+\label{definition:basic-preimage-function}
+    Let $X$ be a set, $Y$ be a topological space, $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, and $P: \mathcal{B} \to 2^X$, then $P$ is a \textbf{basic preimage function} if:
+    \begin{enumerate}[label=(PF\arabic*)]
+        \item $P(\emptyset) = \emptyset$. 
+        \item[(PF2')] For each $\mathcal{U} \subset \mathcal{B}$ and $V \in \mathcal{B}$ with $V \subset \bigcup_{U \in \mathcal{U}}U$, $P(V) \subset \bigcup_{U \in \mathcal{U}}P(U)$.  
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:basic-preimage-function}
+    Let $X$ be a set and $(Y, \topo)$ be a topological space with base $\mathcal{B}$, then:
+    \begin{enumerate}
+        \item For any open preimage function $P: \topo \to 2^X$, $P|_{\mathcal{B}}$ is a basic preimage function.
+        \item For any basic preimage function $p: \mathcal{B} \to 2^X$, there exists a unique open preimage function $P: \topo \to 2^X$ such that $p = P|_{\mathcal{B}}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): For each $U \in \topo$, let $\mathcal{B}(U) = \bracs{V \in \mathcal{B}|V \subset U}$, and
+    \[
+    P: \topo \to 2^X \quad U \mapsto \bigcup_{V \in \mathcal{B}(U)}p(V)
+    \]
+
+    then the only element in $\mathcal{B}$ that can be contained in $\emptyset$ is $\emptyset$ itself, so $P(\emptyset) = \emptyset$. 
+
+    Let $\mathcal{U} \subset \topo$. Since $\bigcup_{U \in \mathcal{U}}\mathcal{B}(U) \subset \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $\bigcup_{U \in \mathcal{U}}P(U) \subset P\paren{\bigcup_{U \in \mathcal{U}}U}$. On the other hand, for each $V \in \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $V \subset \bigcup_{U \in \mathcal{U}}\bigcup_{W \in \mathcal{B}(U)}W$. By (PF2'),
+    \[
+    p(V) \subset \bigcup_{U \in \mathcal{U}}\bigcup_{W \in \mathcal{B}(U)}p(W) = \bigcup_{U \in \mathcal{U}}P(U)
+    \]
+
+    so
+    \[
+    P\paren{\bigcup_{U \in \mathcal{U}}U} = \bigcup_{V \in \mathcal{B}(\bigcup_{U \in \mathcal{U}}U)}p(V) \subset \bigcup_{U \in \mathcal{U}}P(U)
+    \]
+
+    Finally, let $U, V \in \topo$, then since $\mathcal{B}(U \cap V) = \mathcal{B}(U) \cap \mathcal{B}(V)$, $P(U \cap V) = P(U) \cap P(V)$. Therefore $P$ is an open preimage function. 
+    
+\end{proof}
+
+\begin{theorem}
+\label{theorem:open-preimage-function-existence}
+    Let $X$ be a set, $(Y, \fU)$ be a complete Hausdorff uniform space with topology $\topo$, and $P: \topo \to 2^X$ be an open preimage function such that:
+    \begin{enumerate}
+        \item[(S)] For each $x \in X$ and $U \in \fU$, there exists a $U$-small set $V \in \topo$ such that $x \in P(V)$. 
+    \end{enumerate}
+     
+    then there exists a unique $f: X \to Y$ such that $P(U) = f^{-1}(U)$ for all $U \in \topo$.
+\end{theorem}
+\begin{proof}
+    Since $P$ is an open preimage function and $Y$ is Hausdorff, \autoref{proposition:open-preimage-function-gymnastics} implies that such a function is unique if it exists, so it is sufficient to demonstrate existence. 
+
+    For each $x \in X$, let $\fF(x) = \bracs{U \in \topo|x \in P(U)}$, then by (PF1) and (PF3'), $\fF(x)$ is a filter base in $(Y, \fU)$. Moreover, (S) implies that $\fF(x)$ is a Cauchy filter base. Since $Y$ is complete and Hausdorff, there exists a unique $y \in Y$ such that $\fF(x) \to y$. Thus the function in question must be
+    \[
+    f: X \to Y \quad x \mapsto \lim_{y, \fF(x)}y
+    \]
+
+    It remains to verify that $P$ is the open preimage function of $f$. To this end, let $U \in \topo$ and $x \in X$ such that $f(x) \in U$. Since $U \in \cn_Y(f(x))$, there exists a symmetric entourage $V \in \fU$ such that $(V \circ V)(f(x)) \subset U$. By (S), there exists a $V$-small set $W \in \fF(x)$. As $f(x) \in \ol W \subset V \circ W$ and $W$ is $V$-small, $W \subset (V \circ V)(f(x))$. Thus
+    \[
+    x \in P(W) \subset P((V \circ V)(f(x))) \subset P(U)
+    \]
+
+    and $f^{-1}(U) \subset P(U)$. 
+
+    On the other hand, let $x \in P(U)$. Suppose for contradiction that $f(x) \not\in U$. For each $y \in U$, there exists $V_y \in \cn_Y^o(y)$ and $W_y \in \cn_Y^o(f(x))$ such that $V_y \subset U$ and $V_y \cap W_y = \emptyset$. Since $x \in f^{-1}(W_y) \subset P(W_y)$ and $P(V_y) \cap P(W_y) = P(V_y \cap W_y) = \emptyset$, $x \not\in P(V_y)$. By (PF2'), $\bigcup_{y \in U}P(V_y) = P(U)$, so $x \not\in P(U)$, which is a contradiction. Therefore $f^{-1}(U) = P(U)$ for all $U \in \topo$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:basic-preimage-function-existence}
+    Let $X$ be a set, $(Y, \fU)$ be a complete Hausdorff uniform space, $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, and $p: \mathcal{B} \to 2^X$ be a basic preimage function such that:
+    \begin{enumerate}
+        \item[(S')] For each $x \in X$ and $U \in \fU$, there exists a $U$-small set $V \in \mathcal{B}$ such that $x \in P(V)$. 
+    \end{enumerate}
+     
+    then there exists a unique $f: X \to Y$ such that $p(U) = f^{-1}(U)$ for all $U \in \mathcal{B}$.
+\end{corollary}
+\begin{proof}
+    Let $\topo$ be the topology of $Y$. By \autoref{proposition:basic-preimage-function}, $p$ extends to a unique open preimage function $P: \topo \to 2^X$. Since $\mathcal{B} \subset \topo$, \autoref{theorem:open-preimage-function-existence} implies that there exists a unique $f: X \to Y$ such that $f^{-1}(U) = P^{-1}(U)$ for all $U \in \topo$, and in particular for all $U \in \mathcal{B}$. 
+\end{proof}
+
+