Let $X$ be a topological space and $I=\left[0,1\right]$ be the unit interval. Then $C\left(I,X\right)=X^{I}$, which contains all paths in $X$, with compact-open topology, forms a topological space. Are $X^{I}$ and $X$ homotopy equivalent?
My idea is, let $f:X\to X^{I}$ be the continuous mapping sending $x\in X$ to the constant path at $x$. And let $g:X^{I}\to X$ be the continuous mapping sending the path $\gamma :I\to X$ to $\gamma\left(0\right)\in X$. Then we have $g\circ f=id_{X}$. Then it suffices to construct a homotopy from $id_{x^{I}}$ to $f\circ g$.
For any path $\gamma:I\to X$ and $s\in I$, let $\gamma_{s}$ be the path satisfying $\gamma_{s}\left(t\right)=\gamma\left(st\right)$. Then we have $\gamma=\gamma_{1}$.
Then we can construct a mapping $H:I\times X^{I}\to X^{I}:H\left(s,\gamma\right)=\gamma_{s}$.
It is obvious to show that $H\left(0,\cdot\right)=f\circ g,H\left(1,\cdot\right)=id_{X^{I}}$. Visually, it looks like contracting a path to its starting point along itself.
But I can't show that the mapping $H$ is continuous, since the compact-open topology is hard to deal with.
Thanks for help!
As pointed out in the comments, the best thing to use is the hom-tensor adjunction. If you assume your spaces are CW complexes (you can get away with less, but why not do so for the purpose of homotopy theory?), then there is an isomorphism for any space $Y$,
$$ \operatorname{Hom} (Y \times I, X) \cong \operatorname{Hom} (Y, X^I)$$
Taking homotopy classes, since $Y \times I \simeq Y$, there is a natural isomorphism
$$ [Y, X] \simeq [Y, X^I]$$
Hence, by the Yoneda lemma, $X$ and $X^I$ are the same object in the homotopy category.