For topological spaces $X,Y$ let $[X,Y]$ denote the space of continuous maps $X\rightarrow Y$, equipped with the compact-open topology.
According to the Wikipedia page for the compact-open topology,
when $Y$ is locally-compact Hausdorff,
then the composition map
$$
[Y,Z] \times [X,Y] \xrightarrow{ (g,f) \mapsto g \circ f } [X,Z]
$$
is continuous. (Here the LHS has the product topology.)
I was wondering, does anyone have a counter-example showing the above map need not be continuous when $Y$ is not locally-compact Hausdorff?
Best Answer
In any cartesian closed category, the internal hom $[X,Y]$ admits a composition map. Indeed, if $U$ is a generic object then
$$ \begin{align} \text{Hom}(U, [X,Y] \times [Y,Z]) &\cong \text{Hom}(U, [X,Y]) \times \text{Hom}(U, [Y,Z]) \\ &\cong \text{Hom}(U \times X, Y) \times \text{Hom}(U \times Y, Z) \\ &\overset{\star}{\to} \text{Hom}(U \times X, Z) \\ &\cong \text{Hom}(U, [X,Z]) \end{align} $$
Where $\star$ is the map taking $f : U \times X \to Y$ and $g : U \times Y \to Z$ and outputting the map $(u,x) \mapsto g(u,f(u,x)) : U \times X \to Z$
Now, by yoneda, this map must come from a map $[X,Y] \times [Y,Z] \to [X,Z]$.
So to show that $[X,Y]$ equipped with the compact open topology has a composition rule, it suffices to show that it's part of a cartesian closed structure when we restrict to some nice subcategory of $\mathsf{Top}$. But you can find this as a corollary of theorem 4.4 in Booth and Tillotson's Monoidal Closed, Cartesian Closed, and Convenient Categories of Topological Spaces, available here, for instance.
As for your counterexample when $Y$ is not locally compact hausdorff, in the cartesian closed case, the standard example is $[\mathbb{Q}, \mathbb{R}]$.
Here the evaluation map $\mathbb{Q} \times [\mathbb{Q},\mathbb{R}] \to \mathbb{R}$ is discontinuous, as is shown in Internal Hom Objects in the Category of Topological Spaces by Michael Hallam (avaailble here).
Since $X \cong [1, X]$ (notice $1$ is locally compact hausdorff), this shows the natural map (which you can check is the map constructed in the first half of this answer)
$$[1,\mathbb{Q}] \times [\mathbb{Q}, \mathbb{R}] \to [1, \mathbb{R}]$$
is not continuous.
I hope this helps ^_^