Continuity of conjugation actions of Polish groups Let $G$ and $H$ be Polish groups and let $\psi: G \rightarrow H$ be a continuous injective homomorphism such that $\psi(G)$ is normal in $H$.  Then $H$ acts on $G$ by conjugation via $\psi$, in other words for each $h \in H$, we have a unique automorphism $h^*$ of $G$ such that $\psi(h^*(g)) = h\psi(g)h^{-1}$ for all $g \in G$.
Is this action jointly continuous, i.e. is the map $(h,g) \mapsto h^*(g)$ a continuous map from $H \times G$ to $G$?
If not, what about if $G$ and $H$ are locally compact Polish groups?
If something like this is true, a standard reference would be useful.
 A: OK, here is an attempted answer under the assumption that $G$ is locally compact, which can perhaps be refined to give a general answer for Polish groups.  A good reference would still be appreciated though.
Locally compact Polish spaces are Lindelöf (every open cover has a countable subcover).  So if we take a closed subset of $G$, it is a countable union of compact sets.  This means that compact sets generate the Borel $\sigma$-algebra of $G$.  The image of a compact set under $\psi$ is compact, so if $U$ is a Borel subset of $G$, then $\psi(U)$ is Borel in $H$ (and conversely, the preimage of a Borel subset of $H$ is certainly Borel in $G$).  We can see that each element of $H$ induces a Borel automorphism of $G$, and Borel group automorphisms of a Polish group are homeomorphisms.  So $(h,g) \mapsto h^*(g)$ is continuous on $\{h\} \times G$ for each $h \in H$.
Take $K \subseteq G$ compact and $U \subseteq G$ open, and let $B$ be the set of elements of $H$ such that $h\psi(K)h^{-1} \subseteq \psi(U)$.
We see that $U$ can be written as $G \setminus \bigcup_{i < \omega}L_i$ where $L_i$ is compact.  So $h \in B$ if and only if $h\psi(K)h^{-1}$ is a subset of $H \setminus \psi(L_i)$ for every $i < \omega$.  This is a countable conjunction of open conditions on $h$ (because the action of $H$ on itself by conjugation is continuous with respect to the compact-open topology), so $B$ is a Borel set.
We conclude that the map $H \rightarrow Aut(G)$ is Borel-measurable with respect to the (symmetrised) compact-open topology on $Aut(G)$; since $Aut(G)$ is Polish, the map $H \rightarrow Aut(G)$ is therefore continuous; and that is precisely what we need to make $(h,g) \mapsto h^*(g)$ continuous.
