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: For fixed $A,X,Y$ topological spaces (where we can take $A=[0,1]$ for simplicity), 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?

I'm afraid that since there has been no discussion of this idea in the relevant posts below, it may not be interesting nor feasible. Sorry if this is the case.

Relevant: I've seen the following questions:

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

• Another way to view a homotopy is as a map $\widehat{H}:X\rightarrow \mathcal{C}(I,Y)$, that is a collection of paths from $f(x)=H(x,0)=\widehat{H}(x)(0)$ to $g(x)=H(x,1)=\widehat{H}(x)(1)$ indexed by the points of $X$. Really we want all three of the maps $H$, $\tilde{H}$, $\widehat{H}$ to be equally as valid. We end up consider the strongest topology induced by the family of continuous maps $A\times X\rightarrow Y$ where $A$ is any space. And this is simply too strong to be useful. – Tyrone Jun 18 '16 at 13:58
• We would also like to maintain the homeomorphism $\mathcal{C}(\ast,Y)\cong Y$, so this would be equivalent to $Y$ having the final topology for paths in itself, or that $I$ is discrete. – Tyrone Jun 18 '16 at 13:58