# Degree of smooth map (between same-dimensional manifolds) expressed in two ways

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.)

• First, you need to show that the integral $H^n(X) \rightarrow \mathbb{R}$ is an isomorphism (and every $n$-form is closed anyway!). Thus, you know that there is a constant $c$ such that $\int_M{f^*\omega}=c\int_N{\omega}$ for any form $\omega$ on $N$. Now, you should be able to find (by Sard + little things) a contractible subset $U$ in $N$ such that $f^{-1}(U) \rightarrow U$ is a covering map. Then take for $\omega$ a “bump” form supported in $U$. Dec 14, 2021 at 16:52

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.

• Why does it follow from the statement being true for $\omega$ compactly supported in $V$ that the statement is valid for all $\omega$? Jan 14, 2023 at 12:43
• @88888888 since for compact oriented manifold $N$ without boundary of dimensional $n$, every top form $\alpha$ can be written as $\alpha = c\omega + d\beta$ for some $c\in \mathbb R$ and $n-1$ form $\beta$. Also, $\int_N \alpha = c\int_N \omega$ and $\int_M f^*\alpha = c \int_M f^*\omega$ by Stokes theorem. Thus it is sufficient to check that specific $\omega$. Jan 14, 2023 at 12:54