$f^{-1}(L)$ where $L$ is compact.

Question

Let $f: \mathcal{M}\rightarrow \mathcal{N}$ be open, continuous and surjective. $\mathcal{M}$ is locally compact and Hausdorff (there exists a compact neighborhood for every point). $\mathcal{N}$ is only Hausdorff. Prove that for every compact set $L\subset \mathcal{N}$ there is a compact $K\subset\mathcal{M}$ such that $f(K)=L$.

I tried taking $K:=f^{-1}(L)$. Then $f(K)=L\cap \text{Im}(f)=L$ as requested. It is a little more troublesome proving $K$ is compact. I know $L$ is closed ($\mathcal{N}$ is Hausdorff) and therefore $K=f^{-1}(L)$ is also closed. Let us take an arbitrary cover for $K$:

$$K\subset \cup_{i\in I}U_i \Rightarrow f(K)=L\subset\cup_{i=1}^n f(U_i)$$

Here we have used $f$ is open and $L$ is compact. Returning to $\mathcal{M}$ one has:

$$K\subset \cup_{i=1}^n f^{-1}f(U_i)$$

This is not very nice because $f^{-1}f(U_i)$ could be larger then $U_i$. I still haven't used the fact $\mathcal{M}$ is locally compact and $K$ is closed. How do I overcome this difficulty?

Answer 1

$\newcommand{\M}{\mathcal{M}}\newcommand{\N}{\mathcal{N}}$Your answer uses some potentially incorrect set equalities and I felt it was a little unclear. This is my rewriting of the same idea. I also make explicit when each hypothesis is used.

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Take an LCH space $\M$, a Hausdorff space $\N$ and a continuous open surjection $f:\M\to\N$. Fix a nonempty compact $L\subseteq\N$.

As $f$ surjects, $ff^{-1}(L)=L$ and $f^{-1}(L)$ is nonempty. Every $x\in f^{-1}(L)$ has a compact neighbourhood $K_x$. As $f^{-1}(L)\subseteq\bigcup_{x\in f^{-1}(L)}K_x$ it follows that: $$L\subseteq\bigcup_{x\in f^{-1}(L)}f(K_x)$$Since $f$ is open, each $f(K_x)$ is a neighbourhood of $f(x)$ thus $\{f(K_x):x\in f^{-1}(L)\}$ is a cover of $L$ by neighbourhoods, admitting a finite subcover $\{f(K_x):x\in F\}$ for some finite $F\subseteq f^{-1}(L)$.

$\bigcup_{x\in F}K_x$ is compact as a finite union of compact sets. Because $\N$ is Hausdorff and $L$ is compact, $L$ is closed. Because $f$ is continuous, $f^{-1}(L)$ is closed in $\M$. Because $\M$ is also Hausdorff, $\bigcup_{x\in F}K_x$ is closed. Thus $f^{-1}(L)\cap\bigcup_{x\in F}K_x$ is the intersection of two closed sets, so is closed, and it is also compact since it is a closed subset of the compact set $\bigcup_{x\in F}K_x$.

Then define: $$K:=f^{-1}(L)\cap\bigcup_{x\in F}K_x$$

Trivially $f(K)\subseteq L$. $f(K)\supseteq L$ holds because $\bigcup_{x\in F}f(K_x)$ contains $L$, so any $y\in L$ equals $f(z)$ for some $z\in K_x$ for some $x\in F$, and then $z\in f^{-1}(L)$ will also be true so $z\in K$. Therefore $K$ is a compact subset of $\M$ satisfying $f(K)=L$, as desired.

Answer 2

@FShrike's comment got me thinking we should try taking $K_x$ compact neighborhoods in $f^{-1}(L)$ for every $x\in K$ (I am using the notation $f^{-1}(L)=K$). We must have $K\subset \cup_{x\in K} K_x^o$ where the $o$ symbols denotes the interior of the set in $K$'s acquired topology. So $L=f(K)\subset \cup_{x\in K}f(K_x^o)$, where $K_x^o=K\cap A_x$.

$$L\subset \cup_{x\in K} f(K\cap A_{x})=\cup_{x\in K }L \cap f(A_x) $$ Where $A_x$ is an open set in the usual topology for $\mathcal{M}$. We take $L=\cup_{i=1}^nL\cap f(A_{x_i})$ because $L$ is compact and $f$ is open. If we take $\cup_{i=1}^n K\cap K_{x_i}$ we have a compact set because it is a closed subset of $\cup_{i=1}^nK_{x_i}$ which is compact. Furthermore:

$$f(K\cap (\cup_{i=1}^nK_{x_i}))\subset f(K)= L $$

The other inclusion also follows! $$L\subset \cup_{i=1}^nL\cap f(A_{x_i})=\cup_{i=1}^nf(K\cap A_{x_i})\subset \cup_{i=1}^nf(K\cap K_{x_i})= f(\cup_{i=1}^nK\cap K_{x_i} )$$