# Viewing Homotopies as Paths in $\mathcal{C}^0(X,Y)$

When I think intuitively about homotopies, I think about them as paths between two functions. This is more comfortable and suggestive than any categorical talk about "morphisms between morphisms", so I would really like for this intuition to be made precise.

If $X$ and $Y$ are topological spaces and $\phi, \psi: X \to Y$ continuous maps, then a homotopy between $\phi$ and $\psi$ is a continuous map $F: X \times [0,1] \to Y$ such that $F(x,0) = \phi(x)$ and $F(x,1) = \psi(x)$. Alternatively, we may view the homotopy as a one-parameter family of maps $\{f_\alpha \mid \alpha \in [0,1]\}$ where $f_0 = \phi$ and $f_1 = \psi$. From this viewpoint it is straightforward to construct a function $\mathscr F: [0,1] \to \mathcal{C}^0(X,Y)$ given by $\alpha \mapsto f_\alpha$, then $\mathscr F$ is almost a path between $\phi$ and $\psi$: we don't know that $\mathscr F$ is continuous.

The problem is that I don't know what topology ought to be put on $\mathcal{C}^0(X,Y)$. Since the set of continuous functions is subset of $\mathcal{P}(X \times Y)$ we could give it the subspace topology, but that requires a topology on the power set, which my searching shows is not going to be any easier. Even if we could find a topology for either set, we would still need that $\mathscr F$ is continuous with respect to it if and only if $F$ is continuous with respect to the topologies on $X \times [0,1]$ and $Y$.

From Munkres, I have found that

Assuming $X$ is locally compact Hausdorff, we see that $F$ is continuous if and only if $\mathscr F$ is.

Is there any topology for $\mathcal{C}^0(X,Y)$ that can achieve this same result for arbitrary $X$ and $Y$, or is this the best I'm going to get? If so, I can live with it, since $\Bbb R^n$ is locally compact and Hausdorff, but it would be nice to be able to think about all homotopies this way.

• For many situations, paths in $C(X,Y)$ with the compact-open topology are the same as homotopies. I'm not sure what generality this is true in. You might appreciate this MO question. – user98602 Dec 14 '14 at 18:25