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!
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
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 $X,Z$).