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What are the minimal conditions on three topological spaces $X,Y$ and $Z$ such that with the compact-open topology the map

$${(X^Y)}^Z \to X^{Y \times Z}$$

given by taking adjoints is a homeomorpism. The map sends $f: Z \to X^Y$ to $g:Y \times Z \to X$ by the relation $g(y,z)=f(z)(y)$.

This result is known for $Z$ Hausdorff and $Y$ locally compact. I'm interested in a proposition of the form the adjoint construction is a homeomorphism of mapping spaces if and only if some statement regarding the spaces $X$, $Y$ and $Z$. It would also be interesting to see some counterexamples, for example for $Z$ not Hausdorff, etc.

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  • $\begingroup$ Have you looked at the compact-open topology section in a textbook like Munkres? Your question is basically about the fundamental property of the compact-open topology which essentially every textbook treats. $\endgroup$ Sep 25, 2010 at 19:10
  • $\begingroup$ hi,of course. but in nearly every topology book i only can find the "=>",which is enough for most cases but im interested in a <=>. $\endgroup$
    – trew
    Sep 25, 2010 at 20:40
  • $\begingroup$ I do not understand the last ten words of your paragraph. Please reformulate, so as to make this readable. $\endgroup$ Sep 25, 2010 at 20:41
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    $\begingroup$ trew, I believe your question either in the more restricted sense that I wrote up and in the more general sense where you allow any map is an open problem. I might be wrong -- but I posed this question to Stephen Willard when I was an undergraduate and he thought back then it was an open problem. $\endgroup$ Sep 25, 2010 at 21:41
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    $\begingroup$ I'm still having trouble understanding the question. I don't suppose you want the same conditions to apply to all three spaces X, Y, Z? If not, why do you expect a unique answer (you say the minimal conditions)? For example, by slightly strengthening a condition on X one could slightly weaken a condition on Y, so that the two sets of conditions are incomparable. (Finally, may I ask what is the motivation for demanding the compact-open topology, if some other conceptually similar but slightly different topology works?) $\endgroup$
    – Todd Trimble
    Sep 26, 2010 at 11:14

1 Answer 1

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A very closely related question (and maybe the one you meant to ask?) is: which spaces $Y$ in the category of topological spaces and continuous maps are exponentiable, i.e., for which $Y$ does the functor $- \times Y: Top \to Top$ have a right adjoint? A necessary and sufficient condition is that $Y$ is core-compact, as defined at the nLab. See also the references in that article.

There are various ways of defining core-compactness; perhaps the fastest is that the topology is a continuous lattice. It is a slightly weaker condition than local compactness (if local compactness is defined as meaning that every point has a basis of compact neighborhoods), and coincides with local compactness if $Y$ is Hausdorff.

If $Y$ and $Z$ are core-compact, then for every $X$ one can exhibit a canonical homeomorphism

$$(X^Z)^Y \cong X^{Y \times Z}$$

by abstract nonsense (since any two right adjoints to $- \times (Y \times Z)$, in particular $((-)^Y)^Z$ and $(-)^{Y \times Z}$, are canonically naturally isomorphic).

Your question is also interesting when interpreted for locales. See Johnstone's Stone Spaces, where it is shown that a locale is exponentiable if and only if it is locally compact.

If $Y$ is not core-compact, then it is possible to show that there is no exponential $\mathbf{2}^Y$ where $\mathbf{2}$ is Sierpinski space (two points, one open, one closed). In other words, the functor $\hom_{Top}(- \times Y, \mathbf{2})$ is not representable. I once went through the detailed argument (in the case where $Y$ is the space of rational numbers, which I think is illustrative) over at the n-Category Café, see here and the ensuing discussion.

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  • $\begingroup$ Thank you!I didnt know that,but its not the question i meant. $\endgroup$
    – trew
    Sep 26, 2010 at 9:41
  • $\begingroup$ I'd need more clarification then. I'll ask under your question. $\endgroup$
    – Todd Trimble
    Sep 26, 2010 at 10:57

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