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A k-space is a compactly generated Hausdorff topological space. (I used the terminology "k-space" in the question, in order keep the question within the limit of 150 characters.)

Note that under the stated hypotheses in the question, $X$ is automatically Hausdorff and $f$ is automatically a closed map.

Thus, in more basic terms, the question is: If $X$ and $Y$ are topological spaces, with $Y$ being compactly generated and Hausdorff, and $f : X \to Y$ is a proper covering map, then is $X$ necessarily compactly generated?

If the answer is negative, is there a positive answer if $X$ is also assumed to be connected and locally path-connected (both being common assumptions on covering spaces)?

A particular case, in which the answer is positive, is when $Y$ is (Hausdorff and) locally compact. In this case, even fewer hypotheses suffice; for it is well known that if $X$ and $Y$ are Hausdorff, with $Y$ being locally compact, and $f : X \to Y$ is a continuous, proper, surjection, then $X$ is also locally compact. (The converse is also true). Does this result also hold if we replace the hypothesis, "locally compact", with "compactly generated"?

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    $\begingroup$ Here are two suggestions to improve your post. First, do not just leave the question in the title. Put the question in the body of the post as well. Second, explain what a $k$-space is (I did a quick search, $K$-space is a term unsed in functional analysis, and I'm pretty sure that's not what you mean). $\endgroup$
    – Lee Mosher
    Commented Aug 23, 2018 at 1:25
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    $\begingroup$ These suggestions are good, but don't seem to me to be reasons for closing the question (most algebraic topologists would know that $k$-space means compactly generated Hausdorff). It would be good if those who voted to close left a reason - maybe the question is trivial? (I haven't thought about it myself.) $\endgroup$
    – Mark Grant
    Commented Aug 23, 2018 at 8:07
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    $\begingroup$ Doesn't it follow from the covering map being a local homeomorphism? Since an open subset of a k space is also a k space, we get that $X$ is locally a k-space. Now let $A\subset X$ be such that $A\cap K$ is relatively open for any compact $K$. Let $x\in A$ and let $U$ be a neighbourhood of $x$ that is a k space. we have that $A\cap U$ is open in $U$, and so it contains a neighbourhood of $x$ in $X$. Thus, $A$ is open. $\endgroup$
    – erz
    Commented Aug 23, 2018 at 9:53
  • $\begingroup$ @erz Yes, as you have shown, the covering map's property of being a local homeomorphism, ensures that $X$ is compactly generated; and a covering space of a Hausdorff space is automatically Hausdorff itself. $\endgroup$
    – Ian Iscoe
    Commented Aug 25, 2018 at 14:44

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In the question stated the map $f$ has two properties: being proper and being a covering map. In fact, each of these properties by itself is enough to deduce that $X$ is a k-space. The comment by @erz explains that being a covering map is enough: a local $k$-space is a k-space and a covering space of a k-space is a local k-space. I am writing this to document the complementary fact that also the properness of $f$ alone is enough to deduce that $X$ is a k-space.

Claim: A (weakly) Hausdorff topological space admitting a continuous proper map to a (weakly) Hausdorff compactly generated topological space is compactly generated.

The proof of the claim is based on the following easy lemma which I leave as an exercise (but you can have a peek at Lemma 2.13 here).

Lemma: If $g:A\to Y$ is closed and proper and $x_\alpha$ is a net in $A$ such that $g(x_\alpha)$ converges in $Y$ then $x_\alpha$ has a converging subnet.

Proof of the claim: Assume $f:X\to Y$ is a continuous proper map, $X$ is Hausdorff and $Y$ is Hausdorff compactly generated. Fix $A\subset X$. As $X$ is Hausdorff it is clear that for every compact subset $K\subset X$, $A\cap K$ is closed. We will prove the converse. Assume by contradiction that $A$ is not closed and for every compact subset $K\subset X$, $A\cap K$ is closed. Let $x_\alpha$ be a net in $A$ converging to a point not in $A$. Note that $g=f|_A:A\to Y$ is a proper map: for $K\subset Y$, $f^{-1}(K)$ is compact, thus $g^{-1}(K)=f^{-1}(K)\cap A$ is compact too, as it is closed and contained in $f^{-1}(K)$. As $Y$ is a Hausdorff k-space, we get that $g$ is also closed. As $x_\alpha$ has no conveging subnet in $A$, but $g(x_\alpha)$ converges in $Y$ we get a contradiction to the lemma.

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  • $\begingroup$ Along similar lines, there is the following result (see Theorem 3.7.25, on pg. 188 of R. Engelking's book, "General Topology"): "If there exists a perfect mapping f : X --> Y of X onto a k-space Y then X is a k-space." Under the hypotheses of the question, the mapping f is a perfect surjection; so the theorem applies, to obtain the desired conclusion. $\endgroup$
    – Ian Iscoe
    Commented Aug 24, 2018 at 15:02
  • $\begingroup$ @Ian thanks for the perfect comment! $\endgroup$
    – Uri Bader
    Commented Aug 24, 2018 at 19:30
  • $\begingroup$ In the proof of the Claim, don't you also require Y to be Hausdorff, to deduce that f(A) is closed in Y? In detail, to check that the intersection of f(A) with any compact subspace K' of Y, is closed in K', one expresses that intersection as the image under f of the intersection of A with f^{-1}(K'), with f^{-1}(K') being compact in X. Therefore, the latter intersection is closed in f^{-1}(K') and so compact itself. The image under f of that compact intersection is compact and hence closed if Y is Hausdorff. Alternatively, when Y is Hausdorff, one can use the closedness of f. $\endgroup$
    – Ian Iscoe
    Commented Aug 24, 2018 at 21:32
  • $\begingroup$ What if the covering map is not proper, by virtue of having infinite fibres? $\endgroup$
    – David Roberts ♦
    Commented Aug 25, 2018 at 6:09
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    $\begingroup$ @Ian I put the Hausdorff assumption all over the place, just to be sure. In fact, I guess that you don't need to assume it on $X$, only need to assume weakly Hausdorff on $Y$, but I don't care about it that much. Another thing: by your first comment you seem to have known these facts beforehand - so I wonder why did you ask the question in the first place? $\endgroup$
    – Uri Bader
    Commented Aug 25, 2018 at 8:42

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