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Let $X$ be a topological group. $X$ acts freely on itself by left multiplication; this gives us an injective group homomorphism $\Phi: X\rightarrow \operatorname{Homeo} X$. Under what conditions is $\Phi$ also a homeomorphism to its image, in the compact-open topology on $\operatorname{Homeo} X$?

(Is this always true? True if $X$ is locally compact? Is it at least true if $X$ is a Lie group? Etc. Ideally, what's the most general condition and what's the argument?)

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    $\begingroup$ It's obviously true if $X$ is compact Hausdorff. I would venture to say it's true when $X$ is locally compact Hausdorff. $\endgroup$ – Olivier Bégassat Jan 3 '15 at 15:03
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    $\begingroup$ Suppose $X$ is locally compact and locally connected, then by a theorem of Arens (see this answer mathoverflow.net/questions/58690/… by Johannes Ebert) the homeomorphisms of $X$ equipped with the compact open topology form a topological group. Under these conditions, since $\Phi$ is a group homomorphism between topological groups, we only need check openness onto its image on a neighborhood of $1$ $\endgroup$ – Olivier Bégassat Jan 3 '15 at 15:31
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The map $\Phi:G\to C(G,G)$ is always continuous. Indeed, if $K,U$ are compact and open in $G$ respectively, and $g\in\Phi^{-1}([K,U])$, where $[K,U]$ is the set of all continuous maps $f:G\to G$ such that $f(K)\subset U$, then for any $k\in K$ we can choose open nighborhoods $V_k(g)$ and $V(k)$ of $g$ and $k$ respectively such that $V_k(g)\cdot V(k)\subset U$. We then extract a finite open cover $V(k_1),\dots,V(k_n)$ of $K$, and set $$V=\bigcap_{i=1}^n V_{k_i}(g)$$ This is an open neighborhood of $g$, and for any $g'\in V$, $g'K\subset U$, i.e. $V\subset\Phi^{-1}([K,U])$


Suppose $G$ is a locally compact and locally connected topological group, so that the homeomorphisms of $G$ equipped with the compact open topology form a topological group. By a homogeneity argument using the fact that $\mathrm{Homeo}(G)$ is a topological group, it is enough to show that $\Phi$ is an open mapping at $1$.

Let $V$ be an open neighborhood of $1$ in $G$. It is enough to show that there are a compact set $K$ and an open set $U$ such that $$\Phi(1)=\mathrm{id}_G\in\Phi(G)\cap[K,U]\subset\Phi(V)$$ Just choose $U$ and $K$ such that $1\in K\subset U$ and $UU^{-1}\subset V$. Then $\Phi(G)\cap[K,U]$ is open in the induced topology on $\Phi(G)$, and if $\Phi(g)\in\Phi(G)\cap[K,U]$ for some $g\in G$, then $gK\subset U$ so that $g\in UK^{-1}\subset V$.

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