I've been looking at Theorem 2.4.1 from Su Gao's "Invariant Descriptive Set Theory" which deals with the characterization of closed subgroup of $S_\infty$. In particular I'm having difficulties with the (iv) $\Rightarrow$ (i) implication, which states that any Polish group $G$ admitting a countable basis of open sets closed under left multiplication (that is, if $\{ U_n \}_{n \in \omega}$ is the basis, then for each $g \in G$ and $n \in \omega$, $gU_n = U_m$ for some $m$) is a closed subgroup of $S_\infty$.
The proof proceeds as follows: a function $\varphi:G \rightarrow S_\infty$ is defined by letting $\varphi(g)(n) = m$ whenever $gU_n = U_m$ (of course the enumeration for the basis is such that $U_i = U_j \Leftrightarrow i=j$). It is then proved that $\varphi$ is a group isomorphism onto its image, in particular $\varphi(G)$ is a Polish group and thus a closed subgroup of $S_\infty$. My only problem is that the continuity of $\varphi$ is said to be "straightforward" which to me does not seem to be the case after trying to show it directly, by sequences and even just that it is Borel/Baire measurable, which would prove continuity as we're dealing with Polish groups. Am I missing something trivial?
