From 4616ac28612d05ee2f887f2271fa19eeade681c9 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sat, 27 Jun 2026 22:06:50 -0400 Subject: [PATCH] First draft for open preimage functions. --- src/topology/main/index.tex | 1 + src/topology/main/preimage.tex | 111 +++++++++++++++++++++++++++++++++ 2 files changed, 112 insertions(+) create mode 100644 src/topology/main/preimage.tex diff --git a/src/topology/main/index.tex b/src/topology/main/index.tex index c41a869..0b15711 100644 --- a/src/topology/main/index.tex +++ b/src/topology/main/index.tex @@ -26,3 +26,4 @@ \input{./baire.tex} \input{./cube.tex} \input{./compactify.tex} +\input{./preimage.tex} diff --git a/src/topology/main/preimage.tex b/src/topology/main/preimage.tex new file mode 100644 index 0000000..f458cb2 --- /dev/null +++ b/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} + +