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Let $f: X \rightarrow Y$ be a closed surjective continuous map such that $f^{-1}(\{y\})$ is compact for each $y \in Y$. I want to show that if $X$ is regular (Hausdorff) then $Y$ is regular.

I tried to do this using the definition of regularity but got no where. Can you please help?

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Are there any other conditions satisfied by your spaces, Hausdorff, etc? –  gary Jun 21 '11 at 2:24
@gary: Yes, in my definition "regular = regular Hausdorff space", sorry for that. –  user10 Jun 21 '11 at 2:26

1 Answer 1

up vote 2 down vote accepted

The following steps lead to a solution:

(1) Let $y\in Y$ and let $U$ be an open neighborhood of $y$. We wish to find a neighborhood $V$ of $y$ such that $y\in \overline{V}\subseteq U$. (Let us recall that $\overline{V}$ is the closure of $V$.)

(2) Note that $f^{-1}(y)\subseteq f^{-1}(U)$. Prove that if $Z$ is a regular (Hausdorff) space and if $C$ is a compact subset of $Z$ contained in an open subset $W$ of $Z$, then there exists an open neighborhood $N$ of $C$ such that $C\subseteq \overline{N}\subseteq W$. Deduce that there is an open neighborhood $N$ of $f^{-1}(y)$ such that $f^{-1}(y)\subseteq \overline{N}\subseteq f^{-1}(U)$.

(3) Prove that there is a neighborhood $V$ of $y$ in $Y$ such that $f^{-1}(y)\subseteq f^{-1}(V)\subseteq N$. (Hint: use the fact that $f$ is a closed mapping.)

(4) Prove that if $g:A\to B$ is any surjective map and if $S\subseteq B$, then $f(f^{-1}(S))=S$.

(5) Finally, conclude that $V$ is a neighborhood of $y$ and $y\in V\subseteq \overline{V}\subseteq U$.

I hope this helps! Please feel free to ask if you have any questions regarding the above steps.

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There are perfect maps that are not open, so your proof needs work. –  Henno Brandsma Jun 21 '11 at 4:26
You are right; unfortunately, I have not thought about perfect maps for three years and forgot about the "classic example" of a perfect map that is not open. I have edited my post. Thank you for pointing this out! –  Amitesh Datta Jun 21 '11 at 4:53

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