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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

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    $\begingroup$ $\phi$ 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; $\phi^{-1}(a) = \phi(a)$. $\endgroup$ Jan 21 '21 at 18:01
  • $\begingroup$ $\phi$ is continuous and its own inverse, therefore it is an homeomorphism. $\endgroup$
    – lhf
    Jan 22 '21 at 0:49
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The steps in your argument are correct, but I think you've confused yourself by not justifying the key step.

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 $\phi^{-1}(A) = \phi(A) = A^{-1}$.

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  • $\begingroup$ This makes much more sense now. Thanks for pointing this out to me! $\endgroup$
    – Natasha
    Jan 21 '21 at 18:05

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