5
$\begingroup$

Is continuous image of a locally compact space is locally compact?

Let $X$ be locally compact(l.c.). Let $f: X\to Y$ be continuous and surjective.

A space $X$ is locally compact if for each $x\in X$ such that $\exists $ a compact set $V$ such that $x\in U\subset V$ for some open set $U$ containing $x$.

To show that $Y$ is locally compact

My try: Let $y=f(x)\in Y$ and let $V\subset Y$ be an open set containing $y$.then $x\in f^{-1}(V)\subset X$ .As $f$ is continuous then $f^{-1}(V)$ is open in $X$ which is locally compact so there exists a compact set $C\subset X$ such that $x\in f^{-1}(V)\subset C\subset X\implies f(x)\subset V\subset f(C)\subset Y$ Now $f(C)$ being continuous image of a compact set is compact.

Thus proved. But the problem is it has been given that only continuity will not do the map has to be open also.

Please find mistakes in the proof if it exists.

$\endgroup$
  • $\begingroup$ Take $V = Y$ in your proof. What is the problem? $\endgroup$ – Prahlad Vaidyanathan May 18 '15 at 3:38
  • 1
    $\begingroup$ Your definition of locally compact seems to imply that the space is compact, take $U = X$. $\endgroup$ – ronno May 18 '15 at 4:13
  • 11
    $\begingroup$ For a counterexample : Let $X = (\mathbb{Q}, \tau_{\text{discrete}})$ and $Y = (\mathbb{Q},\tau_{\text{usual}})$. Consider $f: X \to Y$ to be the identity map. $X$ is locally compact where as $Y = f(X)$ is not $\endgroup$ – crskhr May 18 '15 at 4:19
  • $\begingroup$ thanks for the example ;@S.C. $\endgroup$ – Learnmore May 18 '15 at 4:22
  • 2
    $\begingroup$ @S.C.: Indeed, you could let $X$ be any non-locally compact space whatsoever, $X_\text{disc}$ the discrete space with the same underlying set, and consider the identity map $X_\text{disc} \to X$. $\endgroup$ – Mike F Jul 26 '17 at 14:59
3
$\begingroup$

I have found an answer ;

Let $y=f(x)\in Y $ then since $X$ is locally compact $x\in X$ has a compact neighborhood i.e there exists an open set $U $ contained in a compact set $V$ i.e. $x\in U\subset V$ then $f(x)\in f(U)\subset f(V)$

Now $f$ being open $f(U)$ is open in $Y$ and continuous image of a compact set being compact implies $f(V)$ is compact

$\endgroup$
2
$\begingroup$

The open set $f^{-1}(V)$ need not be contained in a compact set, so what you have to do is that using locally compactness of $X$ there is a open set $x \in U$ such that there is a compact set $C$ and $U \subset C$. Now take the intersection $U \cap f^{-1}(V)$ and take the image $f( U \cap f^{-1}(V)) \subset f(C)$ now $f$ being open the image is open and $f$ being continuous $f(C)$ is compact. Thus we get locally compactness. The openness condition is required.

$\endgroup$
  • $\begingroup$ Yes, that is exactly the flaw in the proposer's initial attempt. As in the comments, if $X$ is not compact and $V=Y$ there is no compact $C\subset X$ with $f^{-1}V\subset C$. $\endgroup$ – DanielWainfleet Jul 24 '19 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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