Paracompact Hausdorff but not compactly generated?

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I'm sorry to be asking a (possibly) elementary question, but I've run into a problem in point-set topology; I've just read that there exists paracompact Hausdoff spaces which are not compactly generated. I ask the following:

Question: If $X$ is paracompact Hausdorff, is its compactly generated replacement, $k\left(X\right),$ paracompact Hausdorff?

Recall: The inclusion $i:CGH \to Haus$ of compactly generated Hausdorff spaces into Hausdorff spaces has a right adjoint $k,$ which replaces the topology of $X$ with the following topology:

$U \subset X$ is open in $k\left(X\right)$ if and only if for all compact subsets $K \subset X,$ $U \cap K$ is open in $K$.

Another way of describing this topology is that it is the final topology with respect to all maps into $X$ with compact Hausdorff domain. (For the experts, $CGH$ is the mono-coreflective Hull of the category of compact Hausdorff spaces in the category of Hausdorff spaces)

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Touche, but I don't actually think point-set topology is ugly. It was more meant as a joke. Anyway, I'll change it. – David Carchedi May 19 2011 at 5:39
Every compactly generated space is a quotient of a locally compact Hausdorff space. That may help, but not in the naive way. You definitely can't conclude $k(X)$ is paracompact just because it's a quotient of a paracompact space. – David White May 22 2011 at 19:13
Thanks, I'm aware of this result, but I'm not sure how to use it. In fact, this is and if and only if, i.e. it characterizes compactly generated spaces. Moreover, for compactly generated Hausdorff spaces, they are the obvious quotient of the disjoint union of all their compact subsets, and if $X$ is not compactly generated, this quotient is $k\left(X\right).$ This means when $X$ is paracompact Hausdorff, $k\left(X\right)$ is a quotient of a space which is is both locally compact and paracompact Hausdorff. I'm not sure where to go from here. – David Carchedi May 23 2011 at 1:31

1 Answer

(This should be a comment, but my rep is too low.)

It seems that it's certainly Hausdorff, as the topology of $k(X)$ is finer (if $U$ is open in $X$ then $U\cap K$ is open in $K$ for all compacta $K$, by definition of the subspace topology.) So the two separating sets that worked for $X$ still work for $k(X)$.

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Yes, it is indeed Hausdorff; I know that $k$ is a functor $$k:Haus \to CGH,$$ the question is whether or not it is paracompact. – David Carchedi May 18 2011 at 23:47
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