Let $M,N$ be connected compact oriented smooth manifolds (without boundary), both of dimension $n$.
Let $f : M \rightarrow N$ be a smooth map, let $\omega \in \Omega^n(N)$ be a smooth top-degree form.
Could anyone explain why
$$
\int_M f^*\omega = (\mathrm{deg}f)\int_N \omega \;.
$$
Here $\mathrm{deg}f \in \mathbb{Z}$ is the degree of $f : M\rightarrow N$,
computed using the signed orientation-count of a preimage of (any) regular value of $f$. (That this degree is well-defined is proved in differential topology.)
Best Answer
The answer is already briefly described in the comment:
First, one can show that there is $a\in \mathbb R$ such that $$ \int_M f^* \omega = a \int_N \omega$$ for all top form $\omega$ on $N$. This is already non-trivial and is proved here.
Now we see that $a =\mathrm{deg}(f)$. Let $q\in N$ be a regular value of $f$. Then for each $p\in f^{-1}(q)$, there is an open set $U_p $ of $M$ so that $f|_{U_p} \to f(U_p)$ is a diffeomorphism onto an open set $V_p = f(U_p)$. Then $f^{-1}(p)$ is discrete, and thus is finite since $M$ is compact. Write $f^{-1}(q) = \{p_1, \cdots, p_k\}$.
Then one can find a local chart $(V, \psi)$ centered at $q$ and disjoint open subsets $U_1, \cdots, U_k$ of $M$ so that $f^{-1}(V) = U_1\cup\cdots \cup U_k$ and $f|_{U_i} : U_i \to V$ is a diffeomorphism for each $i=1, \cdots, k$.
Let $\omega$ be a bump form (see e.g. here) on $N$, compactly supported in $V$ and $\int_N \omega \neq 0$. Then $f^*\omega$ is supported in $f^{-1}(V)$ (see here) and thus
$$ \int_M f^*\omega = \int_{f^{-1}\ (V)} f^*\omega = \sum_i \int_{U_i} f^*\omega.$$
By the change of variable formula, since each $f|_{U_i}$ is a diffeomorphism,
$$ \int_{U_i} f^*\omega = \pm\int_V \omega $$ where $\pm$ depends if $df_{p_i} :T_{p_i}M \to T_qN$ is orientation preserving or reversing. Thus $$ \int_M f^*\omega = \mathrm{deg}(f) \int_V\omega = \mathrm{deg}(f) \int_N \omega.$$ Since this is true for this $\omega$ and $\int_N \omega \neq 0$, $a=\mathrm{deg}(f)$ and thus the same holds for all top forms.