Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$ and $Y$ be topological spaces, $X$ compact, $f : X \to Y$ continuous. Then the preimage of each compact subset of $Y$ is compact.

With the stipulation that $X$ and $Y$ are metric spaces, this is a theorem in Pugh's Real Mathematical Analysis. The proof uses sequential compactness. Is this theorem true in general (i.e. can it be proved with covering compactness alone)?

share|cite|improve this question
Are your spaces Hausdorff? – egreg Feb 14 '14 at 22:02
It holds if $Y$ is Hausdorff. If it isn't, generally not. – Daniel Fischer Feb 14 '14 at 22:03
Maybe. Does it matter? It's not a homework assignment or anything, some friends and I were just wondering. – Brian Bi Feb 14 '14 at 22:03
Oh okay, can you post a counterexample? – Brian Bi Feb 14 '14 at 22:03
@BrianBi If $Y$ is Hausdorff, a compact subset is closed, so the inverse image is closed in $X$, hence compact. If $Y$ is indiscrete… – egreg Feb 14 '14 at 22:04
up vote 9 down vote accepted

The claim is true if $Y$ is Hausdorff: if $C\subseteq Y$ is compact, then it is closed; therefore $f^{-1}(C)$ is closed in $X,$ hence compact.

For a counterexample, take $Y=X$ with the indiscrete topology and $f$ the identity map. Then every subset of $Y$ is compact, which can be easily arranged for $X$ not to.

share|cite|improve this answer

This is not true in general.

Let $X=Y=[0,1]$. Take $X$ with the usual topology. For $Y$, take the topology $$\tau=\left\{\varnothing,Y,(1/2,1]\right\}.$$ Then $id:x\in X\mapsto x\in Y$ is continuous, but $(1/2,1]=id^{-1}(1/2,1]$ is not compact, although $(1/2,1]$ is compact in $Y$.

On the other hand, if $Y$ is Hausdorff, then every compact of $Y$ is closed, so the inverse image of compact sets is closed, hence compact.

share|cite|improve this answer

A map $f:X\to Y$ is called proper if the preimage of every compact subset is compact. It is called closed if the image of every closed subset is closed.

If $X$ is a compact space and $Y$ is a Hausdorff space, then every continuous $f:X\to Y$ is closed and proper.

Here are some examples where $f$ is not proper:

  • With $X$ compact: Let $X=[0,1]$ and $f=\text{Id}:(X,\tau)\to(X,\sigma)$ where $\tau$ is the Euclidean topology and $\sigma$ is the cofinite topology.
  • With $Y$ Hausdorff: Let $f:\Bbb R\to\{*\}$ be the constant map.
share|cite|improve this answer

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.