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Let $G$ be a group of homeomorphisms of a connected, locally path connected, locally compact metric space $X$. Let Homeo$(X)$ be the group of homeomorphisms of $X$ endowed with the compact open topology. Sets of the form $V(K, U) = \{f \in \text{Homeo} (X) $ | $f(K) \subset U \} $ where $K$ is a compact subset of $X$ and $U$ is an open subset of $X$ form a subbasis for this topology.

While reading a paper I came across the following statement - if $G$ acts effectively on $X$ then we can consider it as a subgroup of Homeo $(X) $ endowed with the compact open topology.

  1. Why is this true? Isn't $G$ automatically a subgroup of Homeo$(X)$?

  2. Does the fact that it acts effectively make it a topological group when given the subspace topology? If so then how do I show that group multiplication (in this case composition) and taking inverses are continuous maps? I tried showing that the inverse image of a subbasic open set is open but got nowhere with it.

Thanks.

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I think the $G$ in the first paragraph and $G$ in the second paragraph are different things. If $G$ is a group of homeomorphisms of a $X$, then of course it is a subgroup of the group of homeomorphisms of $X$, and of course it acts effectively.

On the other hand, if you have any group $G$ acting on $X$ by homeomorphisms, this means that you have a group homomorphism $G\to \operatorname{Homeo}(X)$. Faithfulness of this action corresponds to injectivity of this homomorphism.

In that case, you can use the action (as you could any injective homomorphism), to embed $G$ as a subgroup of $\operatorname{Homeo}(X)$ (in other words, a group of homeomorphisms), and then restrict the compact-open topology of $\operatorname{Homeo}(X)$ to $G$ (seen as a subgroup), or, more formally, you can pull back the topology through the homomorphism (however, in case of a non-injective homomorphism, this will inevitably give you a non-Hausdorff topology).

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Any group action of $G$ on $X$ gives a homomorphism $G \to \text{Homeo}(X)$, by $g \mapsto \varphi_g$, $\varphi_g(x)=g \cdot x$. That the action is effective is the same thing as that this homomorphism is injective.

If $G$ is a topological group then the above homomorphism is continuous if and only if $G$ acts continuously on $X$ in the sense that the map $G \times X \to X$, $(g,x)\mapsto g \cdot x$, is continuous.

Any subgroup of a topological group, when given the subspace topology, is a topological group. Perhaps it will be easier to see in this level of generality.

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