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?
 A: 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.
A: Let be $ [\omega ] \in H^{n}_{c}(M)$ (cohomological group with compact support). By definition of mapping degree, we have $\int_{M} f^{*}(\omega) = deg(f)\int_{N}\omega$ and $\int_{M} g^{*}(\omega) = deg(g)\int_{N}\omega$. Subtracting both equalities we have

$\int_{M} f^{*}(\omega) - g^{*}(\omega) = (deg(f)-deg(g))\int_{N}\omega = (deg(f)-deg(f))\int_{N}\omega = 0$

so 

$\int_{M} (f^{*}- g^{*})(\omega)=0.$

We have to demonstrate $f^{*}=g^{*}$. We know that both $f^{*}(\omega), g^{*}(\omega)$ have compact support, i.e, $\emptyset \neq supp(g^{*}(\omega)) \subset M$ and $\emptyset \neq supp(f^{*}(\omega)) \subset M$, hence, $f^{*}(\omega), g^{*}(\omega)$ are both non trivial, further, $f^{*}(\omega), g^{*}(\omega)$ are smooth differential forms, so, have anyone discontinuos point, therefore, we must have $f^{*}(\omega)- g^{*}(\omega) = 0$. Finally, we have an arbitrary element $\omega$ which implies $f^{*} = g^{*}$, therefore, $f \sim g$.
