Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\mathrm{DISCLAIMER~:~}$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?

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.