When is the suspension-loops adjunction a homeomorphism?

Question

Set up

Let $X=(X,x_0),Y=(Y,y_0),Z=(Z,z_0)$ be pointed topological spaces. Let $\newcommand\Maps{\operatorname{Maps}}\Maps_*(X,Y)$ be the space of continuous pointed maps from $X$ to $Y$ with the compact open topology, and base point the constant map $x\mapsto y_0$. I will use $\newcommand\Top{\operatorname{Top}}\Top_*(X,Y)$ for the underlying set of $\Maps_*(X,Y)$. I will omit the $*$s for the unpointed versions.

Let $\Sigma X \simeq X\wedge S^1$ be the reduced suspension functor, and let $\Omega X\simeq \Maps_*(S^1,X)$ be the loop space functor.

Some facts

For $Y$ locally compact, Hausdorff, $$\Top_*(X\wedge Y,Z)\simeq \Top_*(X,\Maps_*(Y,Z)). $$ In particular, for $Y=S^1$, we have $$\Top_*(\Sigma X,Z) \simeq \Top_*(X,\Omega Z).$$

My question:

When does this bijection of underlying sets induce a homeomorphism $$\Maps_*(\Sigma X,Z)\simeq \Maps_*(X,\Omega Z)?$$

Is this always true, or is there a counterexample when $X$ is not locally compact Hausdorff?

Motivation:

This is (minus constraints on $X$) exercise 18 of Fomenko and Fuchs, Chapter 2.

Some considerations:

If $X$ is locally compact and Hausdorff, this is true.

Proof.

Introduce an auxiliary pointed space $A=(A,a_0)$. Use Yoneda and the smash-hom adjunction: $$ \begin{align} \Top_*(A,\Maps_*(X\wedge S^1,Z)) &\simeq \Top_*(A\wedge(X\wedge S^1),Z)\\ &\simeq \Top_*((A\wedge X)\wedge S^1,Z)\\ &\simeq \Top_*(A\wedge X,\Omega Z)\\ &\simeq \Top_*(A,\Maps_*(X,\Omega Z)).\quad\blacksquare\\ \end{align} $$

Side note: I'm fairly convinced I have a proof that smash product is associative only assuming that the second and third objects are locally compact Hausdorff.

Why I suspect this assumption is necessary:

We can also identify both sides with subsets of $\Top(X\times S^1,Y)\cong \Top(X,\Maps(S^1,Y))$. In Exercise 13, Fomenko and Fuchs ask us to show that the natural map of corresponding spaces is a homeomorphism assuming $X$ and $S^1$ are locally compact and Hausdorff.

I also know that things can go wrong with products/quotients/exponentials when things are not locally compact Hausdorff. See for example these questions. I don't, however, have a counterexample to the claim that $$\Maps(X\times Y,Z)\simeq \Maps(X,\Maps(Y,Z)) $$ whenever $Y$ is locally compact Hausdorff.

Such a counterexample would also be greatly appreciated.

Edit: My proof for associativity of the smash product under these assumptions depended on the smash product $X\wedge S^1$ being locally compact Hausdorff. Unfortunately, I can't see right now why the base point needs to have a compact neighborhood. That said, a proof along the lines indicated in the last section still goes through regardless.

Answer 1

Quite accidentally I've stumbled on a result that basically answers my question.

Proposition A.16 in the appendix of Hatcher is the following:

The natural bijection $$\newcommand\Maps{\operatorname{Maps}}\Maps(X,\Maps(Y,Z)) \simeq \Maps(X\times Y,Z)$$ is a homeomorphism assuming $X$ is Hausdorff and $Y$ is locally compact Hausdorff.

Sketch of proof:

The key point is that $X\times Y$ and $X$ are Hausdorff spaces, so compact subsets of these spaces will be compact Hausdorff, and thus normal. We can use this normality to prove the following two lemmas, which combine to give the result.

Let $M(K,U)$ denote a compact-open subbasic set. Then

1. $M(A\times B, U)$ is a subbasis for $\Maps(X\times Y,Z)$, with $A$ compact in $X$, $B$ compact in $Y$, $U$ open in $Z$.
2. If $X$ is Hausdorff, then for any space $Q$, $M(A,V)$ forms a subbasis for $\Maps(X,Q)$ as $V$ ranges over a subbasis for $Q$, and $A$ ranges over all compact sets in $X$.

Then applying the second result to $Q=\Maps(Y,Z)$, with subbasis $V=M(B,U)$, we find that $M(A,M(B,U))$ is a subbasis for $\Maps(X,\Maps(Y,Z))$.

Since $M(A\times B,U)\leftrightarrow M(A,M(B,U))$ under the natural bijection, this proves that the natural bijection is in fact a homeomorphism. $\blacksquare$

This then answers my question, at least to a point where I am satisfied. (I would still be interested in a counterexample to the conclusion of the proposition when the assumption that $X$ is Hausdorff is dropped.)

To get the desired result, we note that when $X$ is Hausdorff, $Y$ is locally compact Hausdorff, the homeomorphism induces a natural homeomorphism $$\Maps_*(X,\Maps_*(Y,Z)) \simeq \Maps_*(X\wedge Y, Z)$$ using that $\Maps$ applied to a quotient map in the contravariant variable or an embedding in the covariant variable gives an embedding.

Then taking $Y=S^1$, we get that the adjunction $$\Maps_*(\Sigma X, Z)\simeq \Maps_*(X,\Omega Z) $$ is a homeomorphism whenever $X$ is Hausdorff.