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Let $F$ be a topological space such that $G := Homeo(F)$ is a topological group in the compact-open topology. Let $\phi : G \times F \to F$ sending $(g,x)$ to $g(x)$ be the map for the action of $G$ on $F$. I'm wondering if this action is continuous. I can't seem to prove it nor can I find counterxamples, but here is what I have so far.

Let $U$ be open in $F$. I've shown that $\phi^{-1}(U) = \cup_{g \in G} \{g\}\times g^{-1}(U)$. Since $g$ is a homeomorphism, $g^{-1}(U)$ is open in $F$. I'm now struggling to use the compact-open topology to show that the union over $g$ is open, if this is even true.

Or do I need to assume more on $F$ or $G$ to make the group action continuous?

EDIT: In the comments we have that $F$ being locally compact as well is sufficient for $G$ to act continuously. But is it necessary?

Paul Cusson
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  • Related question: https://math.stackexchange.com/questions/2337876/is-the-evaluation-map-yi-times-i-to-y-continuous-for-general-y – Thomas Andrews Sep 23 '20 at 06:49
  • I assume by “compact-open” you mean that $G$ is given the topology inherited from the superset $C(F,F)$ when that set is given the compact-open topology. Then, according to that answer, it is true when $F$ is (strongly?) locally compact. – Thomas Andrews Sep 23 '20 at 06:54
  • Yeah as a subspace of $C(F,F)$. I'm uncertain how that answer relates to my question, perhaps it's a generalization that I don't understand? – Paul Cusson Sep 23 '20 at 07:01
  • The question you are asking is whether $e: G\times F\to F,$ but the linked to question is asking (when $Y=I=F$) if the function $E:C(F,F)\times F\to F$ is continuous, where your $e$ is just the restriction of of $E$ to $G\times F.$ – Thomas Andrews Sep 23 '20 at 07:08
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    @ThomasAndrews that only partially answers the question, i.e. if $F$ is locally compact then the action is continuous. However even if the general $C(F,F)\times F\to F$ action is discontinuous, this doesn't imply that the restriction is discontinuous. Or at least I don't see it. – freakish Sep 23 '20 at 08:16
  • Yeah, there probably aren’t nice necessary and sufficient conditions. $F$ being locally compact in sufficient, but not necessary. – Thomas Andrews Sep 23 '20 at 08:21

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