Let $G$ be a Polish group, that is, a topological group which is separable and metrizable by a complete metric. Let $U,V\subset G$ be open, and consider the set $$ \{ g\in G: gU=V\}.$$
I'm trying to show that this set is open. From standard results on topological groups, it suffices to see that it has the Baire property (and hence it is enough to show that it is Borel), but I cannot show that this is the case.
Any ideas?
In case someone finds this useful, here's a way to do it in case $G$ is second countable (which obviously includes the Polish case).
First, taking complements it is enough to see that $$ \{g\in G: gE=F\} $$ is Borel for $E,F\subset G$ closed. Moreover, since $$gE=F \iff gE\subset F \wedge g^{-1}F\subset E$$ it suffices to study the set $\{g\in G: gE\subset F\}$. Since $G$ is second countable, so is $E$, and in particular $E$ is separable, i.e. there exists a set $\{h_n:n\in\mathbb{N}\}\subset E$ which is dense in $E$. Now, since $F$ is closed,
$$gE\subset F \iff \forall n (gh_n\in F)$$
and this last condition is closed since $Fh_n^{-1}$ is closed. Thus, $\{g\in G:gE\subset F\}$ is closed, and in particular Borel.