Why $h$ has zero topological degree? I am trying to prove that $f,g : M^n \to S^n$, both $C^1$ (indeed just $C^0$ is enough) with the same topological degree are homotopic.
I saw on a book that the trick is as follows:
Take $W = M\times [0,1]$ and define $h(x,0) = f(x),$ and $h(x,1) = g(x).$ Then the map $h : \partial W \to S^n$ has topological degree null, why??
If it is true, then the claim follows from the Hopf theorem, but for the part I stated before, how can I conclude that $h$ has null degree?

The boundary of $M\times [0,1]$ is the disjoint union of two copies $M_1=M\times\{0\}$ and $M_2=M\times\{1\}$ of $M$ with opposite orientation. Thus the degree of $f$ and $g$ are integers with opposite signs.
