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How to solve the following:

Let $X$ be a locally compact, $Y$ Hausdorff space and $f : X\rightarrow Y$ continuous open surjection. Prove that for every compact set $K\subset Y$ exists compact set $C\subset X$ such that $f(C)=K$.

Detailed explanations are welcome.

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    $\begingroup$ The answer by user87690 already seems to be good. $\endgroup$ – Alex Ravsky Dec 19 '13 at 18:01
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We can take the restriction $f ⊇ f': X' = f^{-1}[K] \to K$. Then $X'$ is locally compact and $f'$ is still continuous open surjection. For every $x ∈ X'$ take $U_x$ its compact nbhd. Since $K$ is compact, some finite collection $\{f'[U_x]: x ∈ F\}$ covers $K$. So it is enough to take $C = \bigcup\{U_x: x ∈ F\}$.

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  • $\begingroup$ Why is $f^{-1}[K]$ locally compact? $\endgroup$ – alans Dec 24 '13 at 19:25
  • $\begingroup$ I get it, never mind. $\endgroup$ – alans Dec 24 '13 at 19:38
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Here is what springs to mind, though I am a bit suspicious of my proof.

Suppose we have a compact set $K \subset Y$. For every $y \in K$, we may select an $x \in f^{-1}(y)$, call this element $x(y)$. Define $J = \bigcup_{y \in K}x(y)$. Note that $f(J) = K$.

I claim that $J$ is compact.

Let $\mathcal C = \{U_\alpha\}$ over indexing $\alpha$ be an open cover of $J$. We note that for every $U \in \mathcal C$, $f(U)$ is an open set. Thus, the collection $\{f(U_\alpha)\}$ is an open cover of $K$. Because of the compactness of $K$, we may select a finite subcover $\{f(U_1),\dots,f(U_n)\}$. It follows that $\{U_1, \dots,U_n\}$ is an open cover of $J$ that is finite and a subcover of $\mathcal C$.

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  • $\begingroup$ Note that by choosing $x ∈ f^{-1}(y)$ you actually produce a one-to-one function from $K$ onto $J$ which is continuous since $f$ is open. And so $J$ is compact and $J$ is its $f$-image. So you used only that $f$ is open surjection. It doesn't have to be continous, $X$ doesn't have to be locally compact and $Y$ doesn't have to be Hausdorff. $\endgroup$ – user87690 Dec 13 '13 at 19:23
  • $\begingroup$ @user87690 I am afraid that without providing further specification, we can't be sure that $x(y)$ as I've defined it is a continuous function. For example, for $f(x) = e^{2\pi i x}$, I might define $x(y)$ for numbers of the form $e^{2\pi i p/q}$ (where $p/q$ is in simplest form) as $$x(e^{2 \pi i p/q}) = q + p/q$$ which, I believe, would make $x$ a discontinuous map. $\endgroup$ – Omnomnomnom Dec 15 '13 at 20:24
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    $\begingroup$ Omnomnomnom: Yes you are right, my bad. But then your proof is not correct for the same reason. The fact that $\{f(U_1), …, f(U_n)\}$ covers $K$ does not imply that $\{U_1, …, U_n\}$ covers $J$. $\endgroup$ – user87690 Dec 16 '13 at 17:38
  • $\begingroup$ @user87690 ah, I thought there was a flaw somewhere in the proof. $\endgroup$ – Omnomnomnom Dec 16 '13 at 18:11

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