I am reading the following proof from John Lee's Introduction to Smooth Manifolds. Below is part of the proof on showing that $H$ is embedded if $H$ is a closed Lie subgroup. In the below proof, why does the differential of $\psi$ at $(e,e)$ satisfy $d \psi(X,0)=X$ and $d\psi(0,Y)=Y$ for $X \in T_e V$ , $Y \in T_eS$, and how does this show that $d\psi_{(e,e)}$ is bijective? I cannot understand this line of the proof. I would greatly appreciate any help.
If $\psi$ is the multiplication, to compute ${\rm d}\psi_{(e,e)}(X,0)$, one may take the curve $\alpha(t) = (\alpha_1(t),e)$, where $\alpha_1(0) = e$ and $\alpha_1'(0)=X$. This way: $${\rm d}\psi_{(e,e)}(X,0)= \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \psi(\alpha(t)) = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \alpha_1(t)e = \alpha_1'(0) =X.$$Similarly we have ${\rm d}\psi_{(e,e)}(0,Y) = Y$. It is like computing partial derivatives of functions depending on two variables in Calculus 2. Since you're taking the second entry of $(X,0)$ to be zero and this is based at the point $(e,e)$, it amouns to computing the derivative at the point $e$ evaluated at $X$ of the map $x\mapsto \psi(x,e) = x$, which is the identity.