$G$ a Lie group, $V, S$ submanifolds of $G$ containing $e$, $\psi : V \times S \rightarrow G$; $\psi(v,s)=vs$, then $d\psi(X,0)=X$ and $d\psi(0,Y)=Y$

Question

Okay I couldn't fit this all in the title but here is the full setup:

$G$ a Lie group with $V,S$ submanifolds of $G$ containing the identity element $e$. We are considering the map $\psi: V \times S \rightarrow G$ obtained by restricting the multiplication map on $G$, i.e. $\psi(v,s)=vs$.

I am stuck trying to show that since $\psi(v,e)=v$ for $v \in V$ and $\psi(e,s)=s$ for $s \in S$, then it follows that the differential of $\psi$ at $(e,e)$ satisfies $d\psi(X,0)=X$ and $d\psi(0,Y)=Y$ for $X \in T_{e}V$ and $Y \in T_{e}S$.

I am also confused about how we are identifying $T_{(e,e)}(V \times S)$ with $T_{e}V \oplus T_{e}S$.

This problem arises in the proof of Theorem 7.21 of John Lee's introduction to smooth manifolds book (second edition).

Answer 1

In general, the differential at $(e,e)$ of the multiplication map of a Lie group $m : G \times G \to G$ $$ dm_{(e,e)} : T_eG \oplus T_eG \longrightarrow T_eG $$ satisfy (look here for a proof) $$ dm_{(e,e)}(X,Y) = X+ Y. $$ So by the identification $T_{(v,s)}(V \times S) \simeq T_vV \oplus T_sS$ and previous result in the proof $T_eG = T_eV \oplus T_e S $, the conclusion follows.