Differential forms, pullbacks and determinants I've seen that similar questions have been asked on mathstack, but nevertheless I would appreciate to proof my theorem using local coordinates. Given a differential $n$-form $\omega \in \Omega^n(M)$ I want to show that $$f^\star \omega = \det f'(x) \cdot \omega$$
I know the following:
$$f^\star \omega = \omega_{f(p)} = \sum_{1 \leq i < \ldots < j \leq n}a_{i_1\ldots i_n} dx_{i_1}(f(p))\wedge \ldots \wedge dx_{i_n}(f(p))$$
I can additionally express the determinant using the Leibnitz formula as
$$\det A = \sum_{\sigma \in S_n}{\rm sign}(\sigma)\prod_{i=1}^{n}a_{i\sigma(i)}$$
and I tried to express the coefficients of the differential form with respect to the outer derivative of the respective $n$-form to bring things together, but it simply fails with the indices.
Does anyone already calculated such exercises?
 A: Suppose the map $f$ can be represented in local coordinates as
$$ f : (x_1, \dots, x_n) \mapsto (y_1, \dots, y_n) := (f_1(x_1, \dots, x_n) \ , \ \dots \ , \ f_n(x_1, \dots, x_n)).$$
Then we have
$$ f^\star (dy_i) = \sum_{j = 1}^n\frac{\partial f_i}{\partial x_j} dx_j$$
for every $i \in \{1, \dots, n\}$.
I believe the $n$-form $\omega$ that you are interested in is "unit" volume form in this local basis,
$$ \omega = \ dy_1 \wedge \dots \wedge dy_n.$$
Taking pull-backs, we have
$$  f^{\star}(\omega) = \sum_{j_1 \dots j_n}  \frac{\partial f_1}{\partial x_{j_1}}\dots \frac{\partial f_n}{\partial x_{j_n}} \ dx_{j_1} \wedge \dots  \wedge dx_{j_n} .$$
The only terms that contribute in the sum are the terms where the $j_1, \dots, j_n$ are all distinct, i.e. where there exists a permutation $\sigma \in S_n$ such that $j_1 = \sigma(1), \ \dots, \ j_n = \sigma(n)$. Hence\begin{align}  f^{\star}(\omega) & = \sum_{\sigma \in S_n}  \frac{\partial f_1}{\partial x_{\sigma(1)}}\dots \frac{\partial f_n}{\partial x_{\sigma(n)}} \ dx_{\sigma(1)} \wedge \dots  \wedge dx_{\sigma(n)} \\ & = \sum_{\sigma \in S_n}  \frac{\partial f_1}{\partial x_{\sigma(1)}}\dots \frac{\partial f_n}{\partial x_{\sigma(n)}} \ {\rm sign}(\sigma) \ dx_1 \wedge \dots  \wedge dx_n \\ & = (\det Df ) \ dx_1 \wedge \dots \wedge dx_n.\end{align}
