5.26 Open Preimage Functions
Definition 5.26.1 (Open Preimage Function).label Let $X$ be a set, $(Y, \topo)$ be a topological space, and $P: \topo \to 2^{X}$, then $P$ is an open preimage function if
- (PF1)
$P(\emptyset) = \emptyset$.
- (PF2’)
For each $\mathcal{U}\subset \topo$, $\bigcup_{U \in \mathcal{U}}P(U) = P\paren{\bigcup_{U \in \mathcal{U}}U}$.
- (PF3’)
For each $U, V \in \topo$, $P(U \cap V) = P(U) \cap P(V)$.
Proposition 5.26.2.label Let $X$ be a set, $(Y, \topo)$ be a topological space, and $f: X \to Y$, then:
- (1)
The mapping $U \mapsto f^{-1}(U)$ is an open preimage function.
- (2)
If $Y$ is T1, then for any $g: X \to Y$ with $g^{-1}(U) = f^{-1}(U)$ for all $U \in \topo$, $f = g$.
Proof. (2): Let $y \in Y$, then since $Y$ is T1, $\bracs{x}= \bigcap_{U \in \topo, y \in U}U$, so
Thus for any $x \in X$, $f(x) = y$ if and only if $g(x) = y$, so $f = g$.$\square$
Definition 5.26.3 (Basic Preimage Function).label 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 basic preimage function if:
- (PF1)
$P(\emptyset) = \emptyset$.
- (PF2’)
For each $\mathcal{U}\subset \mathcal{B}$ and $V \in \mathcal{B}$ with $V \subset \bigcup_{U \in \mathcal{U}}U$, $p(V) \subset \bigcup_{U \in \mathcal{U}}p(U)$.
- (PF3’)
For each $U, V \in \mathcal{B}$, $p(U) \cap p(V) \subset \bigcup_{W \in \mathcal{B}, W \subset U \cap V}p(W)$.
Proposition 5.26.4.label Let $X$ be a set, $(Y, \topo)$ be a topological space, and $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, then:
- (1)
For any open preimage function $P: \topo \to 2^{X}$, $P|_{\mathcal{B}}$ is a basic preimage function.
- (2)
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 $p = P|_{\mathcal{B}}$.
Proof. (2): For each $U \in \topo$, let $\mathcal{B}(U) = \bracs{V \in \mathcal{B}|V \subset U}$, and
then $P(\emptyset) = \emptyset$. For any $U \in\topo$, $U = \bigcup_{V \in \mathcal{B}(U)}V$, so if $P$ is an extension of $p$ to $\topo$ as an open preimage function, then $P$ must be unique by (PF2’).
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’),
so
Finally, let $U, V \in \topo$, then since $\mathcal{B}(U \cap V) = \mathcal{B}(U) \cap \mathcal{B}(V)$,
On the other hand, by (PF3’),
so $P(U \cap V) = P(U) \cap P(V)$. Therefore $P$ is an open preimage function.$\square$
Theorem 5.26.5.label 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:
- (S)
For each $x \in X$ and $U \in \fU$, there exists a $U$-small set $V \in \topo$ such that $x \in P(V)$.
then there exists a unique $f: X \to Y$ such that $P(U) = f^{-1}(U)$ for all $U \in \topo$.
Proof. Since $P$ is an open preimage function and $Y$ is Hausdorff, Proposition 5.26.2 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 (S), $\fF(x)$ is non-empty. 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
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, $W \subset (V \circ V)(f(x)) \subset U$. Thus $x \in P(W) \subset P(U)$, 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 $U \in \topo$.$\square$
Corollary 5.26.6.label 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:
- (S’)
For each $x \in X$ and $U \in \fU$, there exists a $U$-small set $V \in \mathcal{B}$ such that $x \in P(V)$.
then there exists a unique $f: X \to Y$ such that $p(U) = f^{-1}(U)$ for all $U \in \mathcal{B}$.
Proof. Let $\topo$ be the topology of $Y$. By Proposition 5.26.4, $p$ extends to a unique open preimage function $P: \topo \to 2^{X}$. Since $\mathcal{B}\subset \topo$, Theorem 5.26.5 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 $U \in \mathcal{B}$.$\square$
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