Topological degree theory Hi all. Is the following statement true? Let $\Omega$ be a bounded open connected region in $R^n$, $f:\bar\Omega\to R^n$, $f\neq 0$ on $\partial\Omega$, $f$ and $\partial\Omega$ smooth, $0$ a regular value of $f$. The degree $deg(f,\Omega,0)$ is the same as the degree of $\partial\Omega\to S^{n-1}$, $x\mapsto f(x)/|f(x)|$... ?
 A: Although the question is elementary, it is an interesting one and I wish to add a remark.
Preamble. A common approach to the Brouwer degree is the axiomatic presentation (one defines its domain as the set $\mathcal{D}$ of "admissible triples" $(f, \Omega, p)$; and to each of these it is attached an integer $\operatorname{deg}(f, \Omega, p)$ in such a way that certain axioms hold, namely "normalization" "continuity" "excision"; one proves that there is one and only one degree map satisfying these axioms). The advantage of an axiomatic presentation, of course, is that even if several constructions are possible, at the end it is not relevant which one we adopt since they lead to the same theory. 
One possible construction of the Brouwer degree is by induction on the dimension: as soon you have the  $\mathbb{Z}$ degree defined on $\mathbb{R}^n$, extend it to $n$ dimensional oriented manifolds, and then to  $\mathbb{R}^{n+1}$ by means of the formula you wrote. You have therefore to check that the axioms for the degree in dimension $n+1$ are satisfied.  
