In this question, Jrrow assumes $\int_N\omega_0\neq 0$ Why is $\deg(f)$ well-defined?
What if $\int_N\omega_0\neq 0$?
My book does not seem to address this explicitly. If $\int_N\omega_0 = 0$, then $\omega_0$ is an exact $n$-form because $H^n(N) \cong \mathbb R$, but then what? If $\int_N\omega_0\neq 0$ is somehow deduced from the assumptions, then please explain how.
My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave. I have a feeling that the diagram here should omit the zero element for each of the 4 vector spaces.
Update: I think I figured it out.
For nonzero integrals only, we have a unique $\deg_{\text{nonzero}}(f)$.
For zero only integrals, we have that any real number could be $\deg{\text{zero}}(f)$ including $\deg_{\text{nonzero}}(f)$.
Therefore, for both nonzero and zero integrals, define $\deg(f) := \deg_{\text{nonzero}}(f)$.
If $M$ is orientable without a boundary, there exists a volune form $\omega_0$ such that $\int_M\omega_0\neq 0$, the map defined on the space of $n$-forms$\Omega^n(M)$ by $I(\omega)=\int_M\omega$ is linear and surjective. The Stokes theorem implies that its kernel contains $d\Omega^{n-1}(M)$ the space of exact $n$-forms, you deduce a surjective linear map $[I]:H^n(M)=\Omega^n(M)/d\Omega^{n-1}(M)\rightarrow \mathbb{R}$ which is an isomorphism since $H^n(M)=\mathbb{R}$. This implies that if $I(\omega)=0$, $[I]([\omega])=0$ where $[\omega]$ is the cohomology class of $\Omega$, since $[I]$ is an isomorphism, $[\Omega]=0$ which is equivalent to saying that $\omega$ is exact.