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

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