# $G$ 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 $d\psi(0,Y)=Y$

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).

By using proposition $3.4$ p. 59 that says $T_{(v,s)}(V \times S) \simeq T_vV \oplus T_sS$ and previous result in the proof $T_eG = T_eV \oplus T_e S$. By regard $\psi(v,e) = v$ and $\psi(e,s)=s$ as identity map in the restricted subset $V\times \{e\}$ and $\{e\} \times S$ respectively, the conclusion follows.
$\textbf{Edit :}$
We can also do this without refering to above proposition. The map $\psi : V \times S \to G$ defined by $\psi(v,s) = vs$. By restricting the domain and codomain of $\psi$ to embedded submanifold $V \times \{e\}$ and $V$ respectively, the map $\psi|_{V \times \{e\}} : V \times \{e\} \to V$ is smooth. Composing with $\iota : V \hookrightarrow V \times \{e\}$, we have identity map $$\psi|_{V \times \{e\}} \circ \iota: V \to V.$$ Since the differential of identity map is also identity map between tangent spaces, we have $$d(\psi|_{V \times \{e\}} \circ \iota)_{e} : T_eV \to T_eV$$ such as for any $X \in T_eV$, $d(\psi|_{V \times \{e\}} \circ \iota)_{e}(X) = X$. By chain rule $$d(\psi|_{V \times \{e\}} \circ \iota)_{e} (X) = d(\psi|_{V \times \{e\}})_{(e,e)} ( d\iota_{e}(X)) = d\psi_{(e,e)}(X,0) = X.$$