I don't understand the proof of the following result in Kechris' "Classical Descriptive Set Theory", pp. $62$
Theorem $(9.14)$
Let $G$ be a group with a topology that is metrizable and Baire, s.t. for each $g\in G$ the function $h\mapsto gh$ is continuous.
Let $X$ be a metrizable space and $(g,x)\mapsto g.x$ an action of $G$ into $X$ which is separately continuous (i.e. the maps $g\mapsto g.x$ for $x\in X$, $x\mapsto g.x$ for $g\in G$ are continuous). Then the action is continuous.
Proof: Fix $(g_0,x_0)\in G\times X$.
By $(8.51)$ the map $(g,x)\mapsto g.x$ la continuous at $(g,x_0)$ for comeager many $g$.
So let $h_0$ be s.t. $(g,x)\mapsto g.x$ is continuous at $(h_0,x_0)$. Since $g.x=(g_0h_0^{-1}).(h_0g_0^{-1}g.x)$, the map $(g,x)\mapsto g.x$ is continuous at $(g_0,x_0)$. QED
Maybe I'm wrong, but I think that the only step I miss is the last row "Since $g.x=(g_0h_0^{-1}).(h_0g_0^{-1}g.x)$, the map $(g,x)\mapsto g.x$ is continuous at $(g_0,x_0)$"
Can someone help me to understand why this is the case?
Thank you in advance for your help.
Best Answer
Choose a neighbourhood $U\times V\subseteq G\times X$ of $(h_0,x_0)$ such that $(g,x)\mapsto g.x$ is continuous on $U\times V$. Let $U'=\{g\in G:h_0g_0^{-1}g\in U\}=g_0h_0^{-1}U$; since left-multiplication is continuous, $U'$ is a neighbourhood of $g_0$. We show that $(g,x)\mapsto g.x$ is continuous on $U'\times V$. We can write $(g,x)\mapsto g.x$ as the composition of three continuous maps $\alpha\circ\beta\circ\gamma$ $$ \require{AMScd} \begin{CD} U'\times V@>{\gamma}>>U\times V@>{\beta}>>X@>{\alpha}>>X \end{CD} $$ where \begin{align*} \gamma(g,x)&=(h_0g_0^{-1}g,x)\\ \beta(g,x)&=g.x\\ \alpha(x)&=(g_0h_0^{-1}).x. \end{align*} $\gamma$ is continuous because it is the left-multiplication by a fixed element of $G$ on the first coordinate and the identity on the second, $\beta$ is continuous because of the choice of $U\times V$, and $\alpha$ is continuous because the action of $G$ on $X$ is separately continuous.