Let TOP be a category of topological spaces and B be an object of TOP. Is there a notion of function space in the comma category TOP/B.
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1$\begingroup$ Can you make "notion of function space" precise? Or do you want a list of the various notions? $\endgroup$– Tom LeinsterCommented Dec 4, 2012 at 17:34
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$\begingroup$ Is the "comma" $TOP/B$ the same thing as the "slice" $TOP/B$? $\endgroup$– Andrej BauerCommented Dec 4, 2012 at 18:02
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$\begingroup$ You say "a" category of topological spaces. There are several ways one might interpret what you mean, but it almost sounds as if you want a full subcategory of $Top$ (ordinary topological spaces) such that every slice $Top/B$ admits function spaces in the sense of being cartesian closed. Or in other words, a full subcategory of $Top$ that is locally cartesian closed. In any case, please explain why you use the word "a". $\endgroup$– Todd TrimbleCommented Dec 4, 2012 at 18:03
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1$\begingroup$ Andrej: presumably. I've heard lots of people say "comma category $C/c$" where "slice category $C/c$" would be more specific. Of course (as you know), a slice category is an example of a comma category. $\endgroup$– Tom LeinsterCommented Dec 4, 2012 at 18:13
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Yes, there is a long history on that subject. It is actually quite subtle point-set topology. The best of the original sources is a series of papers by Peter Booth. A more recent treatment with full details and references is in Section 1.3 of the book "Parametrized homotopy theory" by Johann Sigurdsson and myself. It is available at http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf. [Added] I'll say something about the choice of "a" category $Top$. As usual, we insist that all spaces in sight are $k$-spaces (compactly generated, but with no separation property). We insist that the base space $B$ be weak Hausdorff. We cannot insist that the function space $Map_B(X,Y)$ also be weak Hausdorff, even when the given spaces $X$ and $Y$ over $B$ are weak Hausdorff. Indeed, when that holds, the map $X\longrightarrow B$ is open if and only if $Map_B(X,Y)$ is weak Hausdorff for all $Y\longrightarrow B$, by a result of Gaunce Lewis.
answered Dec 4, 2012 at 17:36