Question about Differential Geometry (Surfaces and Differentiability

Question

I have a doubt about differentiability. Why if I have two regular surface, $S_1$ and $S_2$ $\in \mathbb{R^3}$ and a function $f: \mathbb{R^3} \to \mathbb{R^3}$ such that $f(S_1) \subset S_2$ differentiable, then we can define another function $g=f|_{S_1}$ it's also differentiable? This it's from a proposition of Do Carmo but I can't understand it. And even less can I understand why if it is differentiable we can say that $dg_p(v)=Df(v)$ $\forall p \in S_1$ and $v \in T_p (S_1)$

Answer 1

If you take $i_1:S_1\rightarrow \mathbb{R}^3$ the usual inclusion, then it's differentiable, so $g=f\circ i_1$ is the composition of two smooth maps, then is smooth.

rmk: The inclusion $i_1$ induces a injection $d(i_1)_p:T_p(S_1)\rightarrow \mathbb{R}^3$, so you can talk about $T_p(S_1)\subset \mathbb{R}^3$, and if you look at $g:S_1\rightarrow S_2$, you know that if $\gamma$ is a curve on $S_1$ with $\gamma(0)=p$ and $\gamma'(0)=v\in T_{p}(S_1)$ then $f\circ i_1(\gamma)$ is a curve in $\mathbb{R}^3$ but in fact is a curve in $S_2$ and you can derivative, in this case $dg_{p}(v)=df_{i_1(p)}\circ d(i_1)_p(v)\in \mathbb{R}^3$ but under the identification $T_{f(p)}S_2\subset \mathbb{R}^3$ as we say before with $i_2:S_2\rightarrow \mathbb{R}^3$ we see that $df_{i_1(p)}\circ d(i_1)_p(v)\in T_{f(p)}S_2$.