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$\mathrm{DISCLAIMER~:~}$I am not interested in working with compactly generated spaces.

This post is related to this one : Exponential Law for based spaces. I learned about the exponential law for topological spaces quite some time ago, and I thought I understood it well until I decided to reprove it today as I have been using it lately.

What confuses me is that I 'seem' to have proven it with weaker conditions than those stated in the textbooks. To be precise, I think I have shown that there is a natural homeomorphism $$\mathrm{Map}(X\times Y,Z)\simeq\mathrm{Map}(X,\mathrm{Map}(Y,Z))$$ where $Z$ is any topological space, $X$ is Hausdorff and $Y$ locally compact $no$ $Hausdorff$ $condition$ $required\dots$ In all textbooks I 'm familiar with, none of which feature a proof of the above fact, the extra assumption is made that $Y$ be Hausdorff. The proof I gave is, I think, the one I learnt in Switzer's book (if I remember right) yet I see no need for Hausdorffness in $Y$.

$\mathrm{QUESTION~1:~}$Is $Y$ Hausdorff really necessary?

Also, the reference I am currently using, Algebraic Topology from the Homotopical Viewpoint [Aguilar, Gitler, Prieto, Springer Universitext], exercice $1.3.4$ asks to show that for $X,Y,Z$ topological spaces with $X$ and $Y$ locally compact Hausdorff spaces, composition $$\mathrm{Map}(X,Y)\times\mathrm{Map}(Y,Z)\rightarrow\mathrm{Map}(X,Z),(f,g)\mapsto g\circ f$$ is continuous. Yet I'm pretty sure all you need is for $Y$ to be locally compact$\dots$

$\mathrm{QUESTION~2:~}$ Are all these extra conditions necessary?

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Without Hausdorff, please specify what you mean by locally compact: every point has a base of open sets with compact closure, every point has a base of compact (not open) neighbourhoods, or instead of base you want just one open set with compact closure, or one compact neighbourhood? These are only equivalent under Hausdorff-like conditions... –  Henno Brandsma Jul 4 '11 at 18:17
    
I used following definition : every point has a base of compact neighborhoods. –  Olivier Bégassat Jul 4 '11 at 18:26
    
Ok, so not necessarily compact --> locally compact. –  Henno Brandsma Jul 4 '11 at 18:31
    
Yes, this is no longer true (a priori). –  Olivier Bégassat Jul 4 '11 at 18:36
    
Engelking states this homeomorphism for: any space Y, Z Hausdorff, X locally compact (which includes Hausdorff with him). And for any X,Z Hausdorff, any Y we have an embedding (which need not be onto). Also, the composition map is continuous in the compact open topologies for any X,Z, locally compact Y. –  Henno Brandsma Jul 4 '11 at 18:45

1 Answer 1

up vote 3 down vote accepted

As to Question 2, let $\Sigma$ be the composition map $\mathrm{Map}(X,Y)\times\mathrm{Map}(Y,Z)\rightarrow\mathrm{Map}(X,Z)$.

All functions spaces have the compact-open topology, with as a subbase all sets of the form $\mathrm{M}(C,U) = \{ f: f[C] \subset U \}$, where $C$ is a compact subset of the domain, and $U$ an open subset of the co-domain.

Let $(f,g)$ be a point in $\Sigma^{-1}[\mathrm{M}(C,U)]$, with $C \subset X$ compact and $U \subset Z $ open, and we want to show it's an interior point. We have by definition $(g \circ f)[C] \subset U$, or $f[C] \subset g^{-1}[U]$. As $g^{-1}[U]$ is open, and $f[C]$ is compact, both in $Y$, and if we assume $Y$ is locally compact (in the sense from the comments), we can find for each $y \in f[C]$ a compact neighbourhood $K_y$ that sits inside $g^{-1}[U]$, and so $f[C]$ is covered by finitely many sets of the form $\mathrm{Int}(K_y)$, say $\mathrm{Int}(K_{y_i})$ for $i=1 \ldots n$ and set $W$ to be the finite union of these. Then $W \subset \mathrm{Int}(\cup_{i=1}^n K_{y_i}) \subset K:=\cup_{i=1}^n K_{y_i} \subset g^{-1}[U]$ and so $\Sigma[\mathrm{M}(C,W) \times \mathrm{M}(K,U) ] \subset \mathrm{M}(C,U)$ and so we are done, as $(f,g)$ is in $\mathrm{M}(C,W) \times \mathrm{M}(K,U)$.

This convinces me that indeed we only need local compactness (in a rather strong sense) on $Y$ to show continuity of $\Sigma$, but no Hausdorffness.

share|improve this answer
    
yeah, that was also my proof. –  Olivier Bégassat Jul 4 '11 at 22:08

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