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
Here is a useful application in homotopy theory. It turns out that understanding the homotopy groups of function spaces is greatly facilitated by the use of adjunction maps. This is already very useful, but when we can arrange that these homotopy groups characterise the space entirely, then we can say some really powerful things.
Let $A$ be your favourite finitely generated abelian group and form the Eilenberg-Mac Lane space $K(A,n)$. Let $X$ be a finite, connected CW complex and consider the function space $$Map(X,K(A,n))=K(A,n)^X.$$ It is a theorem of R. Thom that this function space is itself a generalised Eilenberg-Mac Lane space (i.e. a product of Eilenberg-Mac Lane spaces in various dimensions). Here are the details.
To begin we apply a Theorem of J. Milnor, which says that if $X,Y$ are CW complexs and $X$ is finite, then the function space $Map(X,Y)$ is homotopy equivalent to a CW complex. Thus under the above assumptions, $Map(X,K(A,n))$ has CW homotopy type, so to show that it a GEM it will be sufficient to (carefully) check its homotopy groups.
To this end we make use of the fact that CW complexes Hausdorff and finite CW complexes are locally compact. This means that our adjunction maps $E$ are homeomorphisms, and in particular are homotopy equivalences. Using them we have $$\pi_kMap(X,K(A,n))\cong\pi_0\left(Map_*(S^k,Map(X,K(A,n))\right)\cong Map_*(S^k\rtimes X,K(A,n)))\cong \pi_0(Map(X,\Omega^kK(A,n)).$$ Here $Map_*$ denotes the space of pointed maps, $S^k\rtimes X=(S^k\times X)/\ast\times X$ is the half-smash product which we need to introduce to make sense of the adjunctions, and $\Omega^kK(A,n)=Map_*(S^k,K(A,n))$ is the $k$-fold loop space of $K(A,n)$.
Now $\Omega^kK(A,n)\simeq K(A,n-k)$ (this is a point if $k>n$), so the last line in the equation above is just the set of unbased homotopy classes $X\rightarrow K(A,n-k)$. But here we have $$[X,K(A,n-k)]\cong H^{n-k}(X;A).$$ In particular $$\pi_k\left(Map(X,K(A,n))\right)\cong H^{n-k}(X;A).$$
Now the evaluation $ev:Map(X,K(A,n))\times X\rightarrow K(A,n)$ gives a cohomology class in $H^n(Map(X,K(A,n))\times X;A)$, and taking slant products with $H_0(X;H^{n-k}X)\cong H^{n-k}(X)$ we get cohomology classes in $H^n(Map(X,K(A,n));H^{n-k}(X;A))$ (use universal coefficients to identify the coefficient group). These classes are exactly homotopy classes of maps $Map(X,K(A,n))\rightarrow K(H^{n-k}(X;A),k)$. Thom shows that the collection of these maps assemble into a homotopy equivalence $$Map(X,K(A,n))\simeq\prod_{k=0}^nK(H^{n-k}(X,A),k)$$ giving us our claim.