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

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.

• How does your text define locally compact? – Henno Brandsma Jun 27 '17 at 21:13
• @HennoBrandsma the definition that I know is what you call strongly locally compact in the comments below, this is what I was working with. – s.harp Jun 28 '17 at 13:05

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.

• Your proof gives a result that is a little bit more general, namely that if $B$ is locally compact the map $E$ is well defined, meaning $E(f)$ is continuous. The argument I had used to show that evaluation was continuous only needed that $E(\mathrm{id})$ was a continuous function and it did not care about the continuity of $E$ itself. Very interesting! – s.harp Jun 27 '17 at 11:41
• Note that a usual definition of local compactness is that each point has a compact neighbourhood. This implies all compact spaces are (let's call it) weakly locally compact. Adding Hausdorff ensures that it is strongly locally compact in that case, I.e. It has a neighbourhood base of compact sets at every point. This might be reason for the addition of Hausdorff to (maybe weak) local compactness. – Henno Brandsma Jun 27 '17 at 19:07
• @HennoBrandsma I understand what you mean. In my answer you must read it as "strongly locally compact" and some authors (Hatcher for instance) call this "locally compact". This however concerns space $I$ (not $Y$) which of course is often is the notation of the (Hausdorff) unit interval. But the question is: can it be left out that $Y$ is Hausdorff? – drhab Jun 27 '17 at 19:26
• $e$ is continuous if $I$ is strongly locally compact; this is what the proof shows. No separation axioms needed on $I$ or $Y$. We don't need to deduce it from the exponential law at all. But we could as the exponential law holds iff we work with a "core compact" space (see ncatlab.org/nlab/show/exponential+law+for+spaces) which is almost equivalent to strongly locally compact (follow the link locally compact on the previous page for a discussion of several possible definitions for local compactness). – Henno Brandsma Jun 27 '17 at 21:18