To show $V \subset A^{-1}A$ why does it suffice to take $h \in V$ and show $A \cap Ah \neq \emptyset$

Question

I'm reading a theorem on Polish groups...

Theorem: Let $G$ be a topological group and $A \subset G$ a nonmeager subset with the Baire property. Then $A^{-1}A$ contains an open neighborhood of $1_G$.

Proof: Let $U$ be open s.t. $A \Delta U$ is meager. Fix $g \in U$ and let $V$ be open with $1_G \in V$ s.t. $VV^{-1} \subset g^{-1}U$. We claim that $V \subset A^{-1}A$. Let $h \in V$. It suffices to show that $A \cap Ah \neq \emptyset$.

My question is why the last sentence implies $V \subset A^{-1}A$. Any sort of help is greatly appreciated.

Answer 1

Let $x\in A\cap Ah$. Since $x\in Ah$, there is an $a\in A$ such that $x=ah$. Clearly, then, $h=a^{-1}x$, where $a,x\in A$. Thus, $h\in A^{-1}A$. And $h$ was an arbitrary element of $V$, so $V\subseteq A^{-1}A$.