# Topological groups: if $A$ is an open subset of $G$, then $A^{-1}$ is open.

Let $$G$$ be a topological group with open subset $$A$$. I want to prove very thoroughly/precisely that $$A^{-1}:=\{a^{-1}:a\in A\}$$ is also open.

Obviously the key to this proof is the fact that the inversion map $$\phi:G\longrightarrow G$$ is continuous.

$$\phi:G\longrightarrow G$$ is continuous, so since $$A\subseteq G$$ is open in $$G$$, $$\phi^{-1}(A)$$ is open in $$G$$ (by definition of topological continuity).

$$\phi^{-1}(A)$$ consists of elements of the form $$a^{-1}$$, so $$\phi^{-1}(A)=A^{-1}$$. This looks like we are done.

I just want to check a small detail: we take $$A$$ to be in the codomain of $$\phi$$, which of course is $$G$$, but in particular, shouldn't the elements of $$A$$ be of the form "inverses of some elements of $$G$$", so $$A=B^{-1}$$ for some set $$B\subseteq G$$, say, and therefore $$\phi^{-1}(A)=\phi^{-1}(B^{-1})=({B^{-1}})^{-1}=B$$, So we do not necessarily obtain $$A^{-1}$$ back when we apply $$\phi^{-1}$$, unless $$\phi^{-1}$$ is actually an inverse of $$\phi$$, as opposed to a preimage?

I think I might be getting confused here. Could anyone help clarify on this?

Thanks

• $\phi$ is invertible since $G$ is a group, and in fact there's very little point in talking about $\phi^{-1}$ since inversion is an involution; $\phi^{-1}(a) = \phi(a)$. Jan 21 '21 at 18:01
• $\phi$ is continuous and its own inverse, therefore it is an homeomorphism.
– lhf
Jan 22 '21 at 0:49

You say "$$\phi^{-1}(A)$$ consists of elements of the form $$a^{-1}$$" -- why is this? Let's prove it.
Let $$x \in \phi^{-1}(A)$$ be arbitrary. By definition, $$x^{-1} = \phi(x) \in A$$. Then $$x = (x^{-1})^{-1}$$ is the inverse of an element of $$A$$. Since $$x$$ was arbitrary, we have $$\phi^{-1}(A) \subseteq A^{-1}$$. Conversely, for any $$a \in A$$, $$\phi(a^{-1}) = (a^{-1})^{-1} = a \in A$$, so $$a^{-1} \in \phi^{-1}(A)$$. This shows $$A^{-1} \subseteq \phi^{-1}(A)$$, completing the proof.
Here's another way to do this: just note that $$\phi$$ is its own inverse! Thus $$\phi^{-1}(A) = \phi(A) = A^{-1}$$.