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
\[g^{-1}(\bracs{y}) = \bigcap_{\substack{U \in \topo \\ y \in U}}g^{-1}(U) = \bigcap_{\substack{U \in \topo \\ y \in U}}f^{-1}(U) = f^{-1}(\bracs{y})\]
Thus for any $x \in X$, $f(x) = y$ if and only if $g(x) = y$, so $f = g$.$\square$
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