I have a group $G$ with two topologies $\tau, \tau'$ on $G$ that makes it a Polish group (a completely metrizable and separable topological group). I need to show that if $\mathcal{B}(\tau)=\mathcal{B}(\tau') $ , that is they define the same borel sets, then we have $\tau=\tau'$.
I just tried the obvious, I suppose a $U\in \tau$ then $U\in \mathcal{B}(\tau)=\mathcal{B}(\tau')$ but I can't go anywhere from here...
Could you help me?
It suffices to assume that they have the same sets with the Baire property.
It is enough to show that if $e\in U$ is $\tau$-open, then it contains a $\tau'$-open $V\ni e$.
Since $G$ is a $\tau$-topological group, there is a $\tau$-open $W\ni e$ with $W=W^{-1}$ and $WW\subseteq U$. But $W$ is $\tau'$-Borel, so $W$ has the Baire property with respect to $\tau'$, and it is not $\tau'$-meagre (since countably many translates cover $G$), so we can find a nonempty $\tau'$-open $V_0$ such that $W\cap V_0$ is comeagre in $V_0$, and we may assume without loss of generality that $V_0$ is symmetric (replacing it with $V_0\cup V_0^{-1}$ if necessary).
Then you can show that $V_0V_0\subseteq WW\subseteq U$. Since $V_0$ is symmetric, $e\in V_0V_0$ and we are done. Source: Automatic continuity of group homomorphisms.