Let $(U , \varphi)$ and $(V , \psi)$ be two charts on a $n$-dimensional differentiable manifold $M$, with $U \cap V \neq \emptyset$, such that $\varphi = (x_1 , \ldots , x_n)$ and $\psi = (y_1 , \ldots , y_n)$. We have two elements in ${\Lambda}^n(T_pM)$ ($p \in U \cap V$): $$ {(d x_1)}_p \wedge \ldots \wedge {(d x_n)}_p \qquad \mbox{ and } \qquad {(d y_1)}_p \wedge \ldots \wedge {(d y_n)}_p\mbox{.} $$ How can I show that $$ {(d y_1)}_p \wedge \ldots \wedge {(d y_n)}_p = \left(\det d {(\psi \circ {\varphi}^{- 1})}_{\varphi(p)}\right) {(d x_1)}_p \wedge \ldots \wedge {(d x_n)}_p? $$ I have got the equality $$ {(d y_1)}_p \wedge \ldots \wedge {(d y_n)}_p = \lambda(p) {(d x_1)}_p \wedge \ldots \wedge {(d x_n)}_p\mbox{,} $$ where $$ \lambda(p) = \det {\left({(d y_i)}_p {\left(\frac{\partial}{\partial x_j}\right)}_p\right)}_{i , j = 1}^n\mbox{.} $$ I am using the notation ${\left\{{\left(\frac{\partial}{\partial x_i}\right)}_p\right\}}_{i = 1}^n$ basis for $T_pM$ (using the chart $(U , \varphi)$) and ${\{{(d y_i)}_p\}}_{i = 1}^n$ dual basis of ${\left\{{\left(\frac{\partial}{\partial y_i}\right)}_p\right\}}_{i = 1}^n$ for $T_p^*M$ ($= {(T_pM)}^*$).

How can I show that $\lambda(p) = \det d {(\psi \circ {\varphi}^{- 1})}_{\varphi(p)}$?


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