A problem about a equivalence condition of compact subspace

Question

Here are three problems on Marco Manetti's Topology page 78, problem 4.36, 4.37 and 4.38. I guess they are related, so I'd better list all of them here.

4.36 (Compactly-generated spaces) A topological space is called compactly generated if it is Hausdorff and the family of its compact subspaces is an identification cover. Prove that if every point in a Hausdorff space X has a compact neighbourhood then X is compactly generated. ($\{A_{i}\}_{i\in I}$ is an identification cover means it is a cover of $X$ and $U\subset X$ is open if and only if $U \cap A_{i}$ is open in $A_i$ for each $i\in I$)

4.37 (Propermaps)Acontinuousmap $f : Y → Z$ is called proper if the pre-image $f^{−1}(K)$ of any compact set $K ⊂ Z$ is compact. Prove that a Hausdorff space $X$ is compactly generated iff every proper map $f : Y → X$ is closed.

4.38 Let $Z$ be a compact Hausdorff space. Prove that a subset $Y ⊂ Z$ is compact if and only if the projection $X ×Y → X$ is closed, for any space $X$.

I think that problem 4.37 may be useful for 4.38. I can prove 4.36 and if part of 4.37. Here is my proof.

4.36:If $V$ is a compact neighborhood of $x$, $\exists U\subset V$ open. For $K\cap V$ open in $V$. $K \cap V \cap U$ open in $U\cap V =U$. Such $U$ form a open cover of $X$, hence $K$ is open.

4.37:"$\Rightarrow$" For any compact subspace $K\subset X$, $f^{-1}(K)$ is compact. For $A$ closed in $X$, $A\cap f^{-1}(K)$ is closed in $f^{-1}(K)$, hence compact. $f(A\cap f^{-1}(K))=f(A)\cap K$ is compact in $K$, hence closed. Since $K$ is arbitrary, by the definition of identification cover, we have $f(A)$ is closed in $Y$.Thus we have $f$ is a closed map.

I can't prove the "if" part of 4.37. The "only if" part of 4.38 is a basic theorem. For the "if" part, I guess 4.37 is useful in proving it. I found a problem about it in mathematics and there are several answers but none of them use this property. I will be very appreciate if someone can give a proof. Thanks in advance.

Answer 1

4.36 does not seem relevant for the other problems. It's just the definition of compactly generated and an example of a class of spaces that are compactly generated.

To see 4.37, your proof for the forward direction can be more systematically written down: we assume $X$ is compactly generated and assume $f: Y \to X$ is proper.

We then need to show $f$ is also closed, so take $A \subseteq Y$ closed and consider $f[A]\subseteq X$. For any compact subset $K$ of $X$ (by simple set theory, regardless of compactness etc.):

$$f[A]\cap K= f[A \cap f^{-1}[K]]$$

and by properness $f^{-1}[K]$ is compact, so $f^{-1}[K] \cap A$ is also compact in $Z$, being closed in a compact space, and so by continuity its image, $f[A]\cap K$ is compact and thus closed (as $X$ is Hausdorff!). As $K$ was arbitary compact, and the compact subsets form an identification cover, $f[A]$ is indeed closed in $X$.

For the reverse, assume all proper $f$ into $X$ are closed, and we want to see $X$ is compactly generated. So assume $A$ is a subset of $X$ such that $A \cap K$ is compact for any compact subspace $K$ of $X$.

With $X$ we can associate another space $K(X)$: the topological sum of all compact subspaces of $X$ (call them $K_i, i \in I$) under the obvious quotient $q: \bigoplus_{i} K_i \to K(X)$, that identifies "originally identical" points.

Then it's easy to check that the canonical map $i: K(X) \to X$ is a proper map and $A$ considered as a subsaet of $K(X)$ closed by the assumption on $A$, and so $i[A]=A$ is closed as $i$ must be a closed map, completing the proof.