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 $U \in \topo$.$\square$