Your argument for homotopy implying contractability is formally correct, but I am unsure how one would avoid circularity - from where do you get the fact that $[Cf, Z]=0$? The opposite direction is wrong: the isomorphisms of the groups you listed does not imply homotopy equivalence, and the very thread you linked gives an example of why it does not.
But this is not the right way to think about it: there is of course a geometric reason why this is true. If $f:X\to Y$, $g:Y\to X$ realize a homotopy equivalence, there is a homotopy of the identity on $Y$ to the map $fg$, which is of course in the base of the cone $Cf$. This can be extended to a homotopy on all of $Cf$ because the inclusion of the base is a cofibration. Concatenate this homotopy with the retraction to the point of the cone, and we have a contraction of $Cf$.
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
- $M(A\times B, U)$ is a subbasis for $\Maps(X\times Y,Z)$, with $A$ compact in $X$, $B$ compact in $Y$, $U$ open in $Z$.
- 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.
Best Answer
I don't know how explicit we can go, but I'll give it a try. We have to go first through the homotopy-theoretical part.
Since $\{ * \} \subseteq X, \{ * \} \subseteq Y$ are cofibrations, $X \vee Y \subseteq X \times Y$ also is. Let $Z$ be a pointed space and consider the long exact sequence of homotopy for the pair $X \vee Y \subseteq X \times Y$, ie. the sequence
$\ldots \rightarrow [\Sigma ^{2}(X \vee Y), Z] \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X \vee Y), Z] \rightarrow [X \wedge Y, Z] \rightarrow \ldots$,
where $[-,-]$ is the pointed set of homotopy classes of basepoint-preserving maps. Note that for any $n \geq 0$, $\Sigma^{n}(X \vee Y)$ is homeomorphic to $\Sigma^{n}X \vee \Sigma ^{n} Y$. I will not distinguish between the two.
Let $k \geq 1$ and define a map
$\psi ^{k}: \Sigma^{k}(X \times Y) \rightarrow \Sigma^{k}X \vee \Sigma^{k}Y$
$\psi ^{k} = \Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})$,
where $\pi: X \times Y \rightarrow X, Y$ are the projections and $i: X, Y \rightarrow X \vee Y$ are the inclusions. Addition is performed via the suspension structure on $\Sigma^{k}(X \times Y)$, so this is why we require $k \geq 1$. (Observe that even though I denote it by addition this is not necessarily commutative for $k=1$.)
If $j: X \vee Y \hookrightarrow X \times Y$ is the inclusion, then I claim that $\psi ^{k}$ is the left inverse to $\Sigma^{k}j$, ie. $\psi ^{k} \circ \Sigma^{k}j = id_{\Sigma^{k}(X \vee Y)}$. This is important because $\Sigma^{k}j$ are connecting maps in the long exact sequence of homotopy. Indeed, one computes
$\psi ^{k} \circ (\Sigma^{k}j) = (\Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})) \circ \Sigma^{k}j = \Sigma^{k}(i_{X} \pi_{X} j) + \Sigma^{k}(i_{Y} \pi _{Y} j) = \Sigma^{k}(id_{X} \vee const) + \Sigma^{k}(const \vee id_{Y}) \simeq (\Sigma^{k}id_{X} + const) \vee (const + \Sigma^{k}id_{Y}) \simeq \Sigma^{k}id_{X} \vee \Sigma^{k}id_{Y} \simeq id_{\Sigma^{k}X \vee \Sigma^{k}Y}$.
(One can also see this geometrically.) This immediately implies that for all $k \geq 1$ and all $Z$ the $[\Sigma^{k}(X \times Y), Z] \rightarrow [\Sigma^{k}(X \vee Y), Z]$ induced by $j$ is surjective and - by exactness of the long exact sequence - that for all $n \geq 1$ the map $[\Sigma^{n}(X \smash Y), Z] \rightarrow [\Sigma^{n}(X \times Y), Z]$ has zero kernel. In particular, for $k=1$ we have the short exact sequence of groups
$0 \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow 0$
Moreover, the map induced by $\psi^{1}$ splits it and shows that there is a natural isomorphism
$\phi: [\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow [\Sigma(X \times Y), Z]$,
of groups, where the product is only semi-direct, because our groups are not necessarily abelian. This is enough for our purposes, since we also have natural bijections
$[\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y), Z] \times [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), Z]$.
(The second one follows from from the fact that $\vee$ is the direct sum in the category of pointed spaces.) Yoneda lemma establishes that there is an isomorphism
$\theta: \Sigma(X \times Y) \rightarrow _{\simeq} \Sigma(X \smash Y) \vee \Sigma(X) \vee \Sigma(Y) $
in the homotopy category of pointed spaces, ie. a homotopy equivalence that we were after. It takes a little bookkeeping in the above Yoneda-lemma argumentation to see that such map is given by
$\theta = \Sigma(p) + \psi^{i} = \Sigma(p) + \Sigma^{1}(i_{X} \pi_{X}) + \Sigma^{1}(i_{Y} \pi_{Y})$,
where $p: X \times Y \rightarrow X \wedge Y$ is the natural projection. (This is what we get if we start with $id \in [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ and trace it back by all the bijections above to $[\Sigma(X \times Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ - and this is the way to discover the isomorphisms "hidden" by Yoneda lemma.)
I understand that my exposition is far from perfect, but if you would like me to go into more detail over some parts, please comment.