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It is known that the $T_0$ and $T_2$ axioms are not preserved under open, closed and continuous maps (for instance, see here: An example of open closed continuous image of $T_0$-space that is not $T_0$ and here: An example of open closed continuous image of $T_2$-space that is not $T_2$). However, it is not difficult to verify that the Hausdorff property is preserved under perfect functions (closed with compact fibers).

Is it also true that a perfect image of a $T_0$ space is $T_0$ as well? It is not difficult to see that a perfect image of a finite $T_0$ space is indeed $T_0$. What about infinite $T_0$ spaces?

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Yes, it seems to be true.

First, towards a counterexample. Let $f\colon X \to Y$ be a perfect map from a $T_0$ space. If $Y$ is not $T_0$, it contains a two-point indiscrete space $B$, and $f\colon f^{-1}[B] \to B$ is also a perfect map. So if there is a counterexaple at all, there is a counterexample realized by a $T_0$ space $X$ consisting of two compact parts such that $(*)$: every non-empty closed subset intersects both of them. Necessarily $X$ is compact. (In fact, given $(*)$, $X$ is compact if and only if both parts are compact since every open cover of one part is an open cover of $X$.)

However, no compact $T_0$ space satisfying $(*)$ exists. Every compact space contains a minimal non-empty closed subset by Zorn's lemma. And in a $T_0$ space, this minimal closed set has to be a singleton.

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  • $\begingroup$ Reducing the problem to a $T_0$ space with two compact parts such that (*) holds was quite clever, very nice solution! $\endgroup$
    – Peluso
    Commented Mar 12, 2022 at 19:18

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