# Question about proof of n-1 form inducing normal unit vector field

Suppose we have a $$n-1$$ dimensional manifold $$M \subset \mathbb{R}^n$$ and a non-vanishing $$n-1$$ form $$\omega$$ on $$M$$. This implies the existence of a normal unit vector field on $$M$$.

The proof of this goes as follows:

"Let $$x\in M$$, there exists a neighborhood $$U$$ of $$x$$ in $$\mathbb{R}^n$$, a submersion $$f:U\rightarrow \mathbb{R}$$ such that $$M\cap U=f^{-1}(0)$$, $$T_xM=\{u\in T_x\mathbb{R}^n:df_x(u)=0\}$$. Write $$df_x=(\partial f_1(x),...,\partial f_n(x))$$. You can identify $$(\partial f_1(x),...,\partial f_n(x))$$ with a vector $$u_x$$ of $$T_x\mathbb{R}^n$$ such that $$Vect(u_x)$$ is a supplementary space to $$T_xM$$. Let $$\Omega$$ be the canonical volume form $$dx_1\wedge...\wedge dx_n$$, $$(\partial f_1(x)dx_1+...+\partial f_n(x)dx_n)\wedge \omega =c\Omega$$, if $$c>0$$, define $$n(x)={1\over{\|u_x\|}}u_x$$ if $$c<0$$, define $$n(x)=-{1\over{\|u_x\|}}u_x$$. Remark that $$n(x)$$ is well defined in a neighborhood of $$x$$ since it does not depend of the choice of $$f$$. In fact $$u_x$$ is a unit vector orthogonal to $$T_xM$$ relatively to the usual scalar product and it continuously depends of $$x$$."

Anyways, I have a few questions about the proof The first one is: Why is the wedge "$$(\partial f_1(x)dx_1+...+\partial f_n(x)dx_n)\wedge \omega = c\Omega$$" important, how is fundamentally linked to $$n(x)$$, I don't see the importance, and okay $$\omega$$ is non vanishing and there is an $$i$$ such that $$\partial f_i(x)$$ is not zero, but that's no reason to say that the wedge product is not zero, so $$c$$ could still be zero.