# Embedding of topological group in its group of self-homeomorphisms

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?

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 $$\alpha^{-1} = \beta \mid_{L_G}$$.