Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.