Show that a map with some properties is closed

Question

Let X be a topological space and Y hausdorff and local compact.

Let $f:X \rightarrow Y$ be a continuous map such that $f^{-1}(K)$ is compact for all compact sets $K$.

Show that $f$ is a closed map.

I know that the statement would follow if X were a compact Set. This would follow if $f(X)$ were compact. I tried to show this, but it seems impossible.

The other way is to show it directly. So i take a closed subset of X, namely A. There is no other possibility than considering the image $f(A)\subseteq Y$. But how we should continue now? Should we consider the set $f(A)\cap K$ for some compact set K?

Iam thankful for every help!

Answer 1

The fact from your question seems relevant: if $Y$ is locally compact, then $A \subset Y$ is closed iff for all compact $K \subset Y$, $K \cap A$ is compact.

Let $f: X \rightarrow Y$ satisfy the inverse image of compact sets condition. Let $A$ be closed in $X$, and let $K$ be compact in $Y$. Then $f[A] \cap K = f[f^{-1}[K] \cap A]$ (easily shown by proving two inclusions) and the right hand side of this is compact as $f^{-1}[K]$ is compact, so its intersection with $A$ is too (as a closed subset of a compact set is compact) and its image too (as $f$ is continuous). So the condition from the linked question then directly applies to show that $f[A]$ is closed.

Answer 2

This follows from the following lemma:

If $f:X\to Y$ is continuous and each point in $Y$ has a neighborhood $V$ such that the restriction of $f:f^{-1}(V)→V$ is closed, then $f$ is closed.

So if $Y$ is locally compact, what kind of neighborhood could you choose so that the restriction is closed?

Edit: Here is a proof for the lemma above: Let $\mathcal V$ be a cover of $Y$ such that each $y\in Y$ has some $V\in\cal V$ as a neighborhood. It is easy to show that $Y$ is coherent with $\mathcal V$, meaning that a subset $C$ is closed if $C\cap V$ is closed in $V$ for each $V\in\cal V$.
Now let $f:X→ Y$ have the property above. Take a closed $C$ in $X$. It is $f(C)\cap V=f(C\cap f^{-1}(V))$ which is closed in $V$ since $f$ is closed as a map $f^{-1}(V)→V$ for every $V\in\cal V.$ Hence $f(C)$ is closed in $Y.$

Answer 3

Using nets: Suppose $A$ is closed in $X$ and $y$ is the closure of $f[A]$. Then there exists a net $f(x_i)_{i \in I}$ (with all $x_i \in A$) that converges to $y$.

Let $K$ be a compact neighbourhood of $y$. Then $K$ contains a tail of the net $f(x_i)_i$, say all $f(x_i)$ with $i \ge i_0$. Also $f^{-1}[K]$ is compact and contains all those $x_i$, so some subnet $(z_j)_{j \in J}$ exists (with corresponding $s: J \rightarrow I$, such that $z_j = x_s(j)$ and $s[J]$ cofinal in $\{i \in I: i \ge i_0\}$) that converges to some $x_0$ in $X$ (every net has a convergent subnet in a compact space). As $A$ is closed, $x_0 \in A$. And by continuity, $f(z_j) = f(x_s(j))$ converges to $f(x_0)$, which must equal $y$, as limits are unique in Hausdorff spaces and every subnet of a convergent net converges to the same limit. Hence $y = f(x_0) \in f[A]$, which shows that $f[A]$ is closed.