Jacobian of composition

Question

This question comes from the transition function of the tangent bundle of a manifold, but it is more general a question of calculus. Let $U_1,U_2\subset M$ be open sets of a topological set $M$ with $U:=U_1\cap U_2\neq\emptyset$. Let $$ \varphi_1:U_1\rightarrow\mathbb{R^n}\\ \quad \quad \quad \qquad \quad p\mapsto(x^1(p),...,x^n(p)) $$ and similarly $$ \varphi_2:U_2\rightarrow\mathbb{R^n}\\ \quad \quad \quad \qquad \quad p\mapsto(y^1(p),...,y^n(p)) $$ be differentiable functions. How can we prove that the Jacobi matrix of $\varphi_2\circ\varphi_1^{-1}$ is given by $$J(\varphi_2\circ\varphi_1^{-1})=\left[\frac{\partial y^i}{\partial x^j}\right]_{i,j}$$

Answer 1

I've found it, since I really was interested in the transition function between two charts of the tangent bundle. Let $\varphi:U\rightarrow\mathbb{R}^n$ and $\psi:V\rightarrow\mathbb{R}^n$ be two charts of a smooth manifold M, which induces coordinates $(x^1,...,x^n)$ and $(y^1,...,y^n)$ respectively. Then we can induce charts of the tangent bundle by defining $$\Phi:\pi^{-1}(U)\ni v=v^i\frac{\partial}{\partial x^i}|_p\mapsto(x^1(p),...,x^n(p);v^1,...,v^n)\in\mathbb{R}^{2n}$$ and analogously $$\Psi:\pi^{-1}(V)\ni \tilde{v}=\tilde{v}^i\frac{\partial}{\partial y^i}|_p\mapsto(y^1(p),...,y^n(p);\tilde{v}^1,...,\tilde{v}^n)\in\mathbb{R}^{2n}.$$ The inverse of $\Phi$ is easily computed by $$\Phi^{-1}:\mathbb{R}^{2n}\ni(x^1,...,x^n;v^1,...,v^n)\mapsto v=v^i\frac{\partial}{\partial x^i}|_{\varphi^{-1}(x)}\in\pi^{-1}(U).$$ Recall that in the common domain of definition the two tangent basis are related by $$\frac{\partial}{\partial x^i}|_p=\frac{\partial y^j}{\partial x^i}(\varphi(p))\frac{\partial}{\partial y^i}|_p,$$ then the transition map is $$\Psi\circ\Phi^{-1}(x^1,...,x^n;v^1,...,v^n)\mapsto\Psi\left(v^i\frac{\partial}{\partial x^i}|_{\varphi^{-1}(x)}\right)=\Psi\left(v^i\frac{\partial y^j}{\partial x^i}(x)\frac{\partial}{\partial y^j}|_p\right)=\left(y^1(\varphi^{-1}(x)),...,y^n(\varphi^{-1}(x));\frac{\partial y^1}{\partial x^i}(x)v^i,...,\frac{\partial y^n}{\partial x^i}(x)v^i\right)$$