Topology problem - compactness 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.  
 A: 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 $C = \bigcup\{U_x: x ∈ F\}$.
A: 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 $\mathcal C$.
