Showing that $\det(J_f(x))\det(J_{f^{-1}}(x))=1$

Let $$f: U \to V$$ be a diffeomorphism, where $$U$$ and $$V$$ are both subsets of $$\Bbb R^n$$. Show that $$\det(J_f(x))\det(J_{f^{-1}}(x))=1$$ for all $$x \in U.$$

So I have that $$\det(J_f(x)) = \det \begin{bmatrix} \frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex] \frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n} \end{bmatrix}$$ and similarly $$\det(J_{f^{-1}}(x)) = \det \begin{bmatrix} \frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex] \frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n} \end{bmatrix}$$

Now from the fact that $$\det(AB) = \det(A) \det(B)$$ I have that

$$\det \begin{bmatrix} \frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex] \frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n} \end{bmatrix} \cdot \det \begin{bmatrix} \frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex] \frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n} \end{bmatrix} = \det (\begin{bmatrix} \frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex] \frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n} \end{bmatrix} \begin{bmatrix} \frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex] \frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n} \end{bmatrix})$$

But this feels already very messy and I'm not sure how the multiplication turns out... Is there another way I should approach this?

Let $$f$$ map a neighborhood $$U$$ of $$p$$ diffeomorphically onto a neighborhood $$V$$ of $$q:=f(p)$$. Then the map $$g:=f^{-1}:\>V\to U$$ is again differentiable, and we have $$g\bigl(f(x)\bigr)=x\qquad(x\in U)\ .$$ The chain rule then says that $$dg(q)\circ df(p)=d(g\circ f)(p)=d\,{\rm id}_U(p)={\rm id}_{T_p}\ .$$ In terms of matrices this means that $$J_g(q) J_f(p)=I_n\ ,$$ so that $$\det\bigl(J_{f^{-1}}(f(p))\bigr)\det\bigl(J_f(p)\bigr)=1\ .$$ Note that you have a slight typo in the stated formula: The $$J_{f^{-1}}$$ is taken at $$q=f(p)$$, not at $$p$$.
• Could you elaborate on what do you mean by "in terms of matrices"? Isn't $d(g \circ f)(p)$ a matrix already? – user713999 Nov 3 '20 at 19:44
• @Daniel: I denoted the matrix of $df(p)$ with respect to the standard basis by $J_f(p)$, because you had matrices in your question. – Christian Blatter Nov 3 '20 at 20:51