8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How does your text define locally compact? $\endgroup$ – Henno Brandsma Jun 27 '17 at 21:13
  • $\begingroup$ @HennoBrandsma the definition that I know is what you call strongly locally compact in the comments below, this is what I was working with. $\endgroup$ – s.harp Jun 28 '17 at 13:05
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – s.harp Jun 27 '17 at 11:41
  • 2
    $\begingroup$ 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. $\endgroup$ – Henno Brandsma Jun 27 '17 at 19:07
  • $\begingroup$ @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? $\endgroup$ – drhab Jun 27 '17 at 19:26
  • 1
    $\begingroup$ $e$ 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). $\endgroup$ – Henno Brandsma Jun 27 '17 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.