Is the “evaluation map” $Y^{I}\times I\to Y$ continuous for general $Y$?

Question

My question is explicitly the following:

If $Y^{I}$ is given the compact-open topology is the map $Y^{I}\times I\to Y, (\gamma,t)\mapsto \gamma(t)$ continuous even if $Y$ is not Hausdorff?

In the case that $Y$ is Hausdorff one can quickly see that $Y^{I}$ is Hausdorff with the compact-open topology and the statement follows from the exponential law:

Exponential law

If $B$ is locally compact and $C$ is Hausdorff, then the map $$E:(A^{B})^{C}\to A^{B\times C}, \qquad f\mapsto \big(\ (b,c)\mapsto f(c)\,(b)\ \big)$$ is well defined and a homeomorphism, where the spaces of continuous maps have been given the compact-open topology.

The identity function $Y^{I}\to Y^{I}$ is continuous and the above map associates it to the function $Y^{I}\times I\to Y, (\gamma,t)\mapsto \mathrm{id}(\gamma)\,(t)=\gamma(t)$, so the evaluation is continuous.

Some context: The evaluation at $t$ for a fixed $t$ is continuous as a map $Y^{I}\to Y$ regardless of whether or not $Y$ is Hausdorff. This was an exercise and I wonder if this generalisation is always true.

Answer 1

Yes. It is only needed that $I$ is strongly (see the comment of Henno) locally compact.

Let $e$ denote the evaluation function, let $U$ be an open subset of $Y$ and let $e(\gamma,t)=\gamma(t)\in U$.

Since $\gamma$ is continuous and $t$ has arbitrarily small compact neighborhoods a compact neighborhood $K$ of $t$ can be found in $I$ with $\gamma(K)\subseteq U$ or equivalently $\gamma\in M(K,U)$ where $M(K,U)$ is defined as the set $\{f\in Y^I\mid f(K)\subseteq U\}$ and belongs to the subbase of the compact-open topology on $Y^I$.

Then $M(K,U)\times K$ is a neighborhood of $\langle\gamma,t\rangle$ in $Y^I\times I$ that satisfies $e(M(K,U)\times K)\subseteq U$.

This proves that $e$ is continuous at arbitrary $\langle\gamma,t\rangle\in Y^I\times I$, so we conclude that $e$ is continuous.

It was not used in this proof that $Y$ is Hausdorff.