The second part of the answer to the question

When does a topological group embed topologically in its group of homeomorphisms?

made me wary about an argument I have which seems easier and more general.

But let me phrase my question with my own notation. If $G$ is a topological group, we have a map $\varphi \colon G \to \text{Homeo}(G)$ that sends $g$ to the map $L_g$ that multiplies by $g$ on the left. We equip $\text{Homeo}(G)$ with the subspace topology from the space of continuous maps $\text{Map}(G,G)$ with the compact-open topology. If we call $L_G$ the image of $\varphi$, we get a map $\alpha \colon G \to L_G$ which is continuous. The answer to the above question shows that it is a homeomorphism when $G$ is locally compact and locally connected.

I am ok with all that, but I have this argument which seems to show that it is always a homeomorphism. Namely, consider the map $\beta \colon L_G \to G$ that takes $L_g$ to $g = L_g(1)$, which is clearly the inverse of $\alpha$. Let $U$ be an open subset of $G$. Then

$$ \beta^{-1}(U) = \{ f \in L_G \mid f(1) \in U \} = M(\{1\},U) \cap L_G $$

where $M(A,B) = \{ h \in \text{Map}(G,G) \mid h(A) \subseteq B \}$. Since $\{ 1 \}$ is compact and $U$ is open, we have that $M(\{1\},U)$ is open in $\text{Map}(G,G)$ and therefore $\beta^{-1}(U)$ is open in $L_G$.

I do not see any flaw in the argument. Is there anything subtle I am missing?


1 Answer 1


It is correct. You can even define $$\beta : \text{Map}(G,G) \to G, \beta(f) = f(1) .$$ Then $\beta^{-1}(U) = M(\{1\},U)$ which is open in $\text{Map}(G,G)$. Clearly $\beta \mid_{L_G}$ is continuous and $\alpha^{-1} = \beta \mid_{L_G}$.


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