Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be a smooth function. If $a$ is such that $f$ has surjective derivative at all points in $f^{-1}(a)$ then this is an $n-m$ dimensional manifold $X$. I'm trying to show that the following is an orientation form:
$$\alpha = \textrm{det}(\frac{\partial f_i}{\partial x_j})^{-1} dx_1 \wedge ... \wedge dx_{n-m}$$
(This is on a particular chart where the first $m$ columns of $T_xf$ are linearly independent)
Where the determinant is taken over $n-m+1 \le j \le n$ and $1 \le i \le m$.
Clearly this is non vanishing on this particular chart (namely where the last $m$ columns of the matrix $Tf$ give an invertible matrix). Now I'd like to show that if we change coordinates this is still non-vanishing and then we will have a global non-vanishing n-form.
I'm stuck on how to see how $\alpha$ changes under different coordinates. I've tried using the fact that on $X$ we have that $f$ is constant and so for each $i$ we have $\frac{\partial f_i}{x_1}dx_1 + ... + \frac{\partial f_i}{x_n}dx_n = 0$ but how do we know which $i$ to use here to transform the form?
Thanks for any help