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Let $X$ be a compact Hausdorff space. Then if $f: A \to B$ is a map between discrete spaces, the induced map $f^\ast: X^B \to X^A$ is proper.

Question: Are there other classes of map $f: A \to B$ such that $f^\ast: X^B \to X^A$ is proper for every compact Hausdorff space $X$?

Notes:

  • Part of what I find puzzling about the state of affairs is that when $A$ and $B$ are discrete, we don't need to assume that $f$ itself is proper in order to conclude that $f^\ast$ is proper.

  • Of course, in order for the question to make sense, we should probably assume that $A$ and $B$ are exponentiable (locally compact Hausdorff is fine). For all I know, though, it may be the case that there are spaces $A$ which are not exponentiable in general, but for which $X^A$ exists for every compact Hausdorff space; the question would make sense for $A,B$ of this form as well.

  • I'm not certain that restricting $X$ to be compact Hausdorff is the most interesting thing to do, so I'd be interested in this question for other $X$'s as well.

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    $\begingroup$ A small note: if $X^A$ exists for $X$ the Sierpinski space, then $A$ is exponentiable. This doesn't yet address the situation with $X^A$ existing for all compact Hausdorff $X$, but for what interest it has I'm mentioning it. $\endgroup$ Commented Dec 8, 2019 at 2:25
  • $\begingroup$ What do you mean by "proper"? $\endgroup$
    – Wlod AA
    Commented Dec 8, 2019 at 4:54
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    $\begingroup$ @WlodAA I mean that the preimage of every compact set is compact. $\endgroup$ Commented Dec 8, 2019 at 5:32
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    $\begingroup$ My stab in the dark is that local homeomorphisms are approximately what you want. $\endgroup$
    – David Roberts
    Commented Dec 8, 2019 at 8:56
  • $\begingroup$ I find Claim 0 overly strong. Take $X=G$ a compact Lie group, and take $A\to B$ to be $\{0\}\to [0,1]$. Then your map is $ev_0$, evaluation at 0 of the free path space, with fibre the based path space of $G$, which is not compact (I guess this works for $X$ any compact manifold etc), hence $ev_0$ is not proper. $\endgroup$
    – David Roberts
    Commented Dec 8, 2019 at 23:43

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