# Are $X^{I}$ and $X$ homotopy equivalent?

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!

• Perhaps you should use the universal property of $X^I$ as an exponential object. Commented Feb 12, 2023 at 12:15
• Can we just using properties of compact-open topology to show it directly?
– Lier
Commented Feb 12, 2023 at 13:30
• Your $H$ is a map $I\times X^I\rightarrow X^I$. It curries to a map $I\times X^I\times I\rightarrow X,\,(t,\gamma,s)\mapsto H(t,\gamma)(s)$. If this map is continuous, so is the original map. This is a basic property of the compact-open topology and I don't think it's reasonable to avoid it as you'd just reprove it anyhow. Commented Feb 12, 2023 at 13:39

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.
• You don't need any assumption on $X$ or $Y$ for this adjunction to hold. Commented Feb 14, 2023 at 17:21
• @Thorgott you are right, just checked and you need $I$ to be locally compact and Hausdorff (clearly true). I was acting on the side of caution because I always assume everything is a CW complex and forget the conditions. Commented Feb 15, 2023 at 3:54