Is the smash-hom adjunction a homeomorphism? $\newcommand{\top}{\mathsf{Top}}\newcommand{\cg}{\mathsf{CG}}$I will provide references for specific terms and constructions that I'm using, if people ask. In general category theory, we like exponential objects. Unfortunately, the category of spaces, $\top$, does not possess all exponentials. In some instances, however, we can make things work - in $\top$. The Tl;Dr of this question is - although things "work" in the pointed space category $\top_{\ast}$ too, do they work as nicely as they do in $\top$? Specifically, are the natural bijections also homeomorphisms?

Let $X$ be a locally compact Hausdorff (LCH) space and $Y,Z$ any spaces whatsoever. Then it is indeed true that there is a bijection: $$\top(X\times Y,Z)\cong\top(Y,Z^X)$$Where $Z^X$ is the space with point-set $\top(X,Z)$ and the classical compact-open topology. Beautifully and surprisingly, if $Y$ is also an LCH space then, topologising $\top(\cdots)$ with the compact-open topology, the above (natural) bijection is also a homeomorphism.


Let $\cg$ denote the category of all compactly generated spaces, where the following definition is used:

*

*A space $X$ is compactly generated iff. "$A\subseteq X$ is open iff. for all compact Hausdorff spaces $K$, and $u\in\top(K,X)$, $u^{-1}(A)$ is open"

Let $k:\top\to\cg$ be the $k$-ification functor (it is left adjoint to the forgetful $\cg\hookrightarrow\top$). For spaces $X,Y$ understood in $\cg$, their categorical product in $\cg$ will be $X\times_k Y:=k(X\times Y)$. If we let $C_0(X,Y):=\top(X,Y)$ with the compact open topology amended as follows:
A subbasis $(K,U)$ for $C_0(X,Y)$ will be replaced by triplets $(u,K,U)$ which represent: $$\{f\in C_0(X,Y):f(u(K))\subseteq U\}$$For $U\subseteq Y$ open and $u\in\top(K,X)$, for $K$ a compact Hausdorff space.
We then define $C(X,Y)$ as $k(C_0(X,Y))$. It turns out that the following adjunction always works (for $Z\in\cg$) and is a homeomorphism: $$C(X\times Y,Z)\cong C(Y,C(X,Z))$$Which is one reason why $\cg$ is a preferred category.

Now if we decide to do some algebraic topology and work with pointed spaces, some things change. We instead consider the smash-hom adjunction (except in full generality it isn't really an adjunction). To keep the post as brief as possible I'll just say (and provide reference if asked) that:

Assume $a_0$ is the basepoint for a pointed space referred to as $A$. If $X\wedge Y$ denotes the smash product of pointed spaces, then it follows from the above and a quick check that - if $X$ is a LCH space: $$\top_{\ast}(X\wedge Y,Z)\cong\top_{\ast}(Y,Z^X)$$Is a bijection of sets. Likewise, if working in $\cg$ one can define a smash product in the usual way, except that one $k$-ifies the product before the quotient is taken. In that context the above bijection also holds, but $\top_{\ast}$ has to be seen with $k$-ified topology also. Moreover if $Y$ is also LCH (or everything is $\cg$) then there is a homeomorphism: $$\mathcal{M}\cong\top_{\ast}(Y,C(X,Z))$$Of pointed spaces, where: $$\mathcal{M}:=\{f\in\top(X\times Y,Z):f(\{x_0\}\times Y\cup X\times\{y_0\})=\{z_0\}\}\cong\top_{\ast}(X\wedge Y,Z)$$

But is the last bijection a homeomorphism? That would "complete the picture" and show that everything nice about the non-pointed categories falls into nice properties of the pointed categories.
I did attempt this, but it quickly felt far too hard - more of a professional matter, than one for a student new to algebraic / categorical topology. The fundamental difficulty was, given a compact-open subbasic set $(K,U)$ (or $(u,K,U)$) of $\top_{\ast}(X\wedge Y,Z)$, it's not clear how to find a corresponding neighbourhood in $\mathcal{M}$ since $K\subseteq X\wedge Y=X\times Y/(X\vee Y)$ and the preimage of a compact set, out of a quotient space, isn't necessarily compact. This might be a basic stone to stumble over, but I'm not sure how to proceed. I don't even know if it's true.
I would really appreciate references that confirm/deny (under certain conditions) whether or not the smash-hom adjunction is a homeomorphism. Or perhaps it's an open (uninteresting?) problem. Either way, I think I'm in a little out of my depth here, so I'm asking for help!
 A: Here is a categorical approach to the problem.
The smash product is defined as the pushout
$\require{AMScd}$
\begin{CD}
X\sqcup Y @>{}>> X\times Y\\
@VVV @VVV\\
\ast @>{}>> X\wedge Y
\end{CD}
The pointed internal hom is the pullback
$\require{AMScd}$
\begin{CD}
[X,Y]_* @>{}>> [X,Y]\\
@VVV @VVV\\
[\ast,\ast] @>{}>> [\ast,Y]
\end{CD}
where the maps are induced by pre-composing with the basepoints of $X$ and $Y$.
By plugging in a definition for the pullback in Top, you can figure out that $[X,Y]_*$ is indeed the set of basepoint preserving maps and its topology is generated by the compact-open topology in $[X,Y]$.
These constructions exist for any closed monoidal category $\mathcal C$ with pullbacks/pushouts.
Then it's basically folklore$^*$ that in this case

*

*$(\mathcal C_*, \wedge, \ast\sqcup \ast)$ is a monoidal category;


*$[X,Y]_*$ is an internal hom for $\mathcal C_*$.
These two facts imply that when enriching $\mathcal C_*$ over itself we will obtain
$$
[X\wedge Y,Z]_*
\cong
[X,[Y,Z]_*]_*
$$
which is what we wanted for $\mathcal C = \mathsf{Top}$.
* You can find an elementary proof in Section 4 of the paper Permutative categories, multicategories, and algebraic K-theory by Elmendorf and Mandell.
