# Topology on $\mathcal{C}(X,Y)$ to work with homotopy.

We know that the compact open topology on $\mathcal{C}(X,Y)$ is a good choice for topology on the set of continuous maps, but this seems really efficient, both naively and with respect to existence of good theorems, when $X$ is a locally compact Hausdorff topological space.

For instance, given a map $H:A \times X \to Y$, then, if $X$ is locally compact Hausdorff, we can assure that

$H$ is continuous if and only if the induced map $\widetilde{H}: A \to \mathcal{C}(X,Y)$ given by $\widetilde{H}(a)=H(a, \cdot)$ is continuous.

Now, in my intuition, I prefer to see homotopies as the maps $\widetilde{H}$, when $A=[0,1]$. However my point of view is not formally justified, given that there is an inconsistency regarding to what are homotopies in the two points of view, since the "iff" statement above need not hold on general spaces.

Therefore, I was thinking, and this is my question:

Q: What happens when instead of considering the compact open topology, we consider the final topology on $\mathcal{C}(X,Y)$, with respect to the family $\{\widetilde{H_{\lambda}}\}_{\lambda \in \Lambda}$ of induced maps coming from continuous maps $H_{\lambda}: A \times X \to Y$? Is this manageable/interesting/has this been done?

Relevant: I've seen the following questions:

but none of them addresses the suggestion made here, which is the main interest of this question.

• The final topology is with $A$ varying over all spaces (this is how I interpreted it in my answer), or only with $A = [0,1]$?. – Goa'uld Jun 15 '16 at 13:14
• @Goa'uld No, only with $A=[0,1]$. I think taking it to "vary over all spaces" can also have some set-theoretic issues, but I should have made it clear. thank you. Regarding your answer, thank you very much for the attention, but I am interested exactly on the cases for which $X$ is not locally compact Hausdorff. – Aloizio Macedo Jun 16 '16 at 22:53