I'm trying to understand a sentence in a theorem about Polish groups. I'll write the proof all the way up to the sentence that I'm having trouble with.
Theorem: Let $G,H$ be Polish groups and $\phi: G \rightarrow H$ be a group homomorphism. If $\phi$ is Baire measurable, then $\phi$ is continuous.
Proof: Let $U \subset H$ be open and let $g \in \phi^{-1}(U)$. Let $V \subset H$ be open s.t. $1_H \in V$ and $V^{-1}V \subset \phi(g)^{-1}U$.
I'm having trouble with how to find such a $V$. I know that since $\phi(g) \in U$ we have that $1_H \in \phi(g)^{-1}U$.
We have $\phi(g) \in U$. Therefore $\phi(g)^{-1}U$ is an open neighbourhood of $1_H$. In every topological group $T$, the map $\psi \colon (x,y) \mapsto x^{-1}y$ is continuous, in particular it is continuous at $(1_T,1_T)$. That means for every neighbourhood $N$ of $1_T$, there are neighbourhoods $A,\, B$ of $1_T$ such that
$$\psi (A\times B) = A^{-1}B \subset N.$$
Then $C = A \cap B$ is a neighbourhood of $1_T$ with $C^{-1}C \subset N$.
Take $N = \phi(g)^{-1}U$.