I have a doubt on compactly generated spaces. The majority of the question is here for context, you can skip through the definitions and the facts mentioned to the question, if you're familiar with them. Thanks for any clarify.
Let $X$ be a topological space.
- $X$ is said weak Hausdorff if every compact subset is closed;
- a subset $S\subset X$ is compactly closed if, for every continuous map $K\to X$ from a compact space $K$, $f^{-1}(S)\subset K$ is closed;
- $X$ is said a $k$-space if every compactly closed subset is closed.
Let $kX$ be the topological space defined as follows:
- the underlying set is (the same of) $X$;
- a subset $S\subset kX$ is closed if and only if $S$ is compactly closed in $X$.
Fact 1. The identity map on the underlying sets $i:kX\to X$ is continuous, and $S\subset kX$ is closed if and only if $i(S)\subset X$ is compactly closed.
Fact 2. $kX$ is a $k$-space.
Let $C\subset kX$ be compactly closed, and let $f:K\to X$ be a continuous map from a compact space $K$. $f$ factors as $i\circ f'$, for a (unique) continuous map $f':K\to kX$: the definition on the underlying sets is evident, and it is continuous because, for any closed $S\subset kX$, we have $f'^{-1}(S)=f^{-1}(i(S))$, which is closed as $i(S)$ is compactly closed. Now $f^{-1}(i(C))=f'^{-1}(C)$, which is closed as $C$ is compactly closed, yielding the thesis.
Fact 3. If $X$ is weak Hausdorff, also $kX$ is.
Let $C\subset kX$ be compact. $i(C)$ is compact too, hence closed by hypothesis. So $C=i^{-1}(i(C))$ is closed.
Fact 4. If $X$ is weak Hausdorff, if $C\subset X$ is compact also $i^{-1}(C)$ is.
We show that the continuous map $i':i^{-1}(C)\to C$ induced by $i$ is a homeomorphism: it is sufficient to prove that $i'$ is closed. Let $S'\subset i^{-1}(C)$ be closed, so that $S'=S\cap i^{-1}(C)$ for some closed $S\subset kX$. Then $i'(S')=i(S)\cap C$, which is closed in $C$ since $i(S)$ is compactly closed.
Question. Is it necessary that $X$ is weak Hausdorff for Fact 4 to be true?
It was assumed on the text I'm reading, but it doesn't seem necessary. The point is that, in any space $X$, holds this: ($*$) if $S\subset X$ is compactly closed, then $S\cap C\subset C$ is closed for every compact $C\subset X$; if $X$ is weak Hausdorff, we know that, if $S\subset X$ satisfies ($*$), then $S$ is compactly closed, but we don't need this direction in the proof of Fact 4, I think.
Indeed, the weak Hausdorff hypothesis is unnecessary in Fact 4. Here's how I would think about it. The topology of $kX$ is the finest topology that makes every continuous map from a compact space to $X$ still continuous as a map to $kX$. In particular, if $C\subseteq X$ is compact, the inclusion map $C\to kX$ is still continuous, so the topology of $kX$ is the same as the topology of $X$ when restricted to $C$.