The homotopy type of $\mathrm{Maps}\left[X,Y\right]$ depends only on the homotopy types of $X,Y$ Let $X,Y$ be two topological spaces and $\mathrm{Maps}\left[X,Y\right]$ be the collection of continuous maps $X\to Y$ with the compact-open topology. Is the homotopy type of $\mathrm{Maps}\left[X,Y\right]$ only dependent on the homotopy types of $X,Y$. That is, is I choose two spaces $X\simeq X'$, $Y\simeq Y'$ which are homotopy equivalent to $X,Y$, is it true that $\mathrm{Maps}\left[X,Y\right]\simeq \mathrm{Maps}\left[X',Y'\right]$. What if I replace homotopy equivalences by weak homotopy equivalences, if the statement true or false now? If it is false in either of the cases, are there good conditions to put on $X,Y$ such that it becomes true?
 A: It is true that $Map(X,Y)$ depends only on $X,Y$ through their homotopy types. There are a couple of tricks to make the proof work. See tom Dieck's book Algebraic Topology pg. 40.
Actually the result follows more generally from the fact that a homotopy $H_t:X\times I\rightarrow Y$, $f\simeq g$, induces homotopies
$$Map(H_t,Y):Map(X',Y)\times I\rightarrow Map(X,Y),\qquad Map(Y,H_t):Map(Y,X)\times I\rightarrow Map(Y,X')$$
between the induced maps $f^*\simeq g^*$ and $f_*\simeq g_*$. These homotopies are defined the obvious way, and the trick lies in verifying their continuity.
It is not true that the same result holds if homotopy equivalence is replaced by weak equivalence. Let $\mathbb{S}$ be the digital circle. This is a finite topological space with four points which is a quotient of $S^1$. The quotient projection $q:S^1\rightarrow \mathbb{S}$ is a weak equivalence. It is not a homotopy equivalence, however, since the only maps $\mathbb{S}\rightarrow S^1$ are the constant maps. It follows that
$$q^*:Map(\mathbb{S},S^1)\rightarrow Map(S^1,S^1)$$
is not even bijective on path components.
With respect to covariant functorality things are a little better. I can prove the following result when working in any convenient category of spaces. I need at least all objects to be exponentiable. To be concrete let us assume compactly generated spaces. Of course $Map(X,Y)$ does not now carry the compact-open topology, but rather its compactly generated replacement. I do not know if the same statement can be make in all of $Top$, but I strongly believe that it fails.

The following statements are equivalent for a given map $f:X\rightarrow Y$.

*

*$f$ is a weak equivalence.

*For any CW complex $K$, the induced map $f_*:Map(K,X)\rightarrow Map(K,Y)$ is a weak equivalence.

*For any CW complex $K$, the induced map $f_*:[K,X]_0\rightarrow [K,Y]_0$ is bijective.


