1

Let $M\subset\mathbb{R}^n$ and $N\subset\mathbb{R}^m$ be two manifolds and $f:U\rightarrow\mathbb{R}^m$ a function of class $C^1$ defined in an open set $U\supset M$ and such that $f(M)\subset N$.

To prove: $Df(a)(T_M(a))\subset T_N(f(a))$

It's clear that $Df(a)$ is differentiable in $U$ and continuous in $\mathscr{L}(\mathbb{R}^n,\mathbb{R}^m)$, so I can write $Df(a)$ as $Df(a)(x)=J_f(a).x^t$, where $x^t$ is the column vector $x$ and $J_f(a)$ is the jacobian matrix of $f$ in $a$, but I don't know how to use that.

If I take an $u\in T_M(a)$ I know that there exists a curve $\gamma:]-\delta,\delta[\rightarrow M$ such that $\gamma(0)=a$ and $\gamma'(0)=u$.

How do I prove that there exists a curve $\beta:]\varepsilon,\varepsilon[\rightarrow N$ such that $\beta(0)=f(a)$ and $\beta'(0)=Df(a)(u)$? Is there any other way to prove that $Df(a)(u)\in T_N(f(a))$?

Thank you very much.

Alejandro
  • 50
  • 1
    Try $\beta=f\circ\gamma$. – Martín-Blas Pérez Pinilla Feb 19 '14 at 10:34
  • Then we have a curve $\beta:]\delta,\delta[\rightarrow f(M)\subset N$ such that $\beta(0)=f(\gamma(0))=f(a)$ and $\beta'(0)=D(f\circ\gamma)(0)=Df(\gamma(0))\circ \gamma'(0)=Df(a)(u)$.

    Is the chain rule well applied, isn't it? I didn't manage to get to that solution, but now it seems really easy. Thank you very much, @Martín-Blas.

     – Alejandro Feb 19 '14 at 10:53

0 Answers0