The path components of $Map(X,Y)$ is one to one corresponding to $[X,Y]$, the set of homotopy classes between $X$ and $Y$.

Question

Show that The path components of $Map(X,Y)$, equipped with compact-open topology with a subbasis $$ \mathcal{O}_{K,U}:=\{f\in Map(X,Y): f(K)\subseteq U\},$$ is one to one corresponding to $[X,Y]$, the set of homotopy classes between $X$ and $Y$.

My attempt: Let $f,g\in Map(X,Y)$. We need to show $f,g$ are in the same path component iff $f\simeq g$.

$\Rightarrow $: Suppose $\phi:I\rightarrow Map(X,Y)$ is a path from $f$ to $g$. We define $$ \begin{aligned} H:X\times I&\rightarrow Y\\ (x,t)&\mapsto \phi(t)(x)\end{aligned}$$ We need to show $H$ is continuous. Let $V$ be an open subset of $Y$, we check $H^{-1}(V)$.

Step1: $(x,t)\in H^{-1}(V)$ means $$ \phi(t)(x)\in U \Leftrightarrow \phi(t)\in \mathcal{O}_{\{x\},V}\Leftrightarrow t\in \phi^{-1}( \mathcal{O}_{\{x\},V})$$ Since $\phi$ is continuous, $\phi^{-1}( \mathcal{O}_{\{x\},V}))$ is open, therefore we can choose an open interval $(a_x,b_x)$, such that $$t\in (a_x,b_x)\subseteq \phi^{-1}(\mathcal{O}_{\{x\},U})$$ So we have shown that $$(x,t)\in H^{-1}(V)\Rightarrow \{x\}\times (a_x,b_x)\subseteq H^{-1}(V)$$

Step2: $(x,t)\in H^{-1}(V)$ means $$x\in (\phi(t))^{-1}(V)$$ Therefore we can find a neighborhood of $x$, say $U_t$, such that $$(x,t)\in H^{-1}(V)\Rightarrow U_t\times \{t\}\subseteq H^{-1}(V)$$

I have no idea how to proceed. It seems that I have exhausted all the properties of $\phi$. Can I conclude that $H^{-1}(V)$ is open just from step1 and step2?

By the way, the other direction is much easier.

THANKS!

Answer 1

What you need here is known as the exponential law for function spaces. There are many questions in this forum about it. Also have a look at any topology textbook dealing with function spaces.

For topological spaces $Z, Y$ let us write $Z^Y$ for the set $Maps(Y,Z)$ endowed with the compact-open topology. This associates to $Y,Z$ the function space $Z^Y$.

We can define a function $$E : Z^{X \times Y} \to (Z^Y)^X, E(f)(x)(y) = f(x,y).$$ It is easy to verify that this is well-defined which means that

1. All functions $E(f)(x) : Y \to Z$ are continuous.
2. $E(f) : X \to Z^Y$ is continuous.

The function $E$ is obviously injective, but not always surjective. It is surjective if $Y$ is locally compact. The exponential law states a bit more; actually $E$ is a homeomorphism if $X$ is Hausdorff and $Y$ is locally compact. But this fact is irrelevant here.

Let us apply this for $X = I$. We get an injection $$E : Z^{I \times Y} \to (Z^Y)^I$$ which is a bijection for locally compact $Y$.

The maps $f : I \times Y \to Z$ are nothing else than homotopies written in a bit unusual form, and the maps $g : I \to Z^Y$ are paths in the function space $Z^Y$.

Using $E$, we see that homotopic maps $Y \to Z$ always lie in the same path component of $Z^Y$. But only for locally compact $Y$ we can conclude that $E$ induces a bijection between the set of homotopy classes $[Y,Z]$ and the set of path components of $Z^Y$.

In $\operatorname{Hom}(X \times Z, Y) \cong \operatorname{Hom}(X, \operatorname{Map}(Z, Y))$ is not true in $\textbf{Top}$ you will find an example of a space $Y$ (which is of course not locally compact) for which $E : Z^{X \times Y} \to (Z^Y)^X$ is not bijective (at least not for all pairs $X,Z$).