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I am trying to solve the following problem.

Let $X$ be locally compact Hausdorff and $Y$ be Hausdorff.

(a) If $f: X \to Y$ is continuous and open map then show that $f(X)$ is locally compact.

(b) If $f: X \to Y$ is onto continuous and open map then show that for any compact subset $K$ of $Y$ there is a compact subset $C$ of $X$ such that $f(C)=K$.

I have solved (a), for (b) I thought I could solve it straightforwardly by using the fact that $f(f^{-1}(K))=K$ since $f$ is onto and show that $C=f^{-1}(K)$ is compact in $X$ but as shown in the below picture I have confronted a problem. So I've thought about using the result in (a) but my efforts have been unsuccessful so far. Can anyone help me?

enter image description here

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  • $\begingroup$ $f^{-1}(K)$ is in general not compact. But if you start with a covering $\{ U_\alpha : \alpha \in A\}$ of $f^{-1}(K)$ such that $\overline{U_\alpha}$ is compact for all $\alpha$, what do you get? $\endgroup$
    – Daniel Fischer
    Nov 28, 2014 at 21:27
  • $\begingroup$ Hmm... I don't really see anything yet can you explain a bit more? $\endgroup$
    – nomadicmathematician
    Nov 28, 2014 at 21:52
  • $\begingroup$ How can I come up with such a covering in the first place? $\endgroup$
    – nomadicmathematician
    Nov 28, 2014 at 21:53
  • $\begingroup$ $X$ is locally compact, so for every $x\in C$, let $U_x$ be a relatively compact open neighbourhood of $x$. $\endgroup$
    – Daniel Fischer
    Nov 28, 2014 at 21:55
  • $\begingroup$ Okay I see how I can come up with such a covering but I still don't see where you intend to take me can you help me out? $\endgroup$
    – nomadicmathematician
    Nov 28, 2014 at 22:01

1 Answer 1

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For every $x\in f^{-1}(K)$, consider an open neighborhood $U_x$ of $x$, with compact closure. Then $f(U_x)$ is open and the family $\{f(U_x):x\in f^{-1}(K)\}$ is an open cover of $K$. Then $$ K\subseteq f(U_{x_1})\cup f(U_{x_2})\cup\dots\cup f(U_{x_n}) $$ for some $x_1,\dots,x_n\in f^{-1}(K)$. Take $$ C=(\overline{U_{x_1}}\cup \overline{U_{x_2}}\cup\dots\cup\overline{U_{x_n}})\cap f^{-1}(K) $$

Then $C$ is compact and $f(C)\subseteq f(f^{-1}(K))=K$. If $y\in K$, then $y=f(x)$ with $x\in U_{x_i}$ for some $i$, $1\le i\le n$. So…

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