# Why is $\deg(f)$ well-defined?

If $M$ and $N$ are boundaryless, compact, connected, oriented $n$-manifolds, and $f:M\to N$ is smooth, then if $\omega_0$ is a $n$-form on $N$ and $\int_N\omega_0\neq 0$, there is a number $a$ such that $\int_M f^*\omega_0=a\int_N\omega_0$.

In fact, if $\omega$ is any $n$-form on $N$, then $\int_M f^*\omega=a\int_N\omega$. How do we know it's the same $a$ for any $\omega$?

## 2 Answers

First, it is a well known theorem that if $M$ is a connected, orientable $n$-manifold without boundary, then $H^n_c(M)\simeq\mathbb{R}$. Since your manifolds are compact, $H^n_c(M)=H^n(M)$, the latter being the usual de Rham cohomology group. This is a $1$-dimensional $\mathbb{R}$-vector space.

By Stokes' theorem, $\displaystyle\int_N\omega_0\neq 0$ implies $\omega_0$ is not exact, so its cohomology class $[\omega_0]$ spans $H^n(N)$. Let $\omega\in\Omega^n(N)$. Then $[\omega]=c\cdot[\omega_0]$ for some $c$, or $\omega-c\omega_0=d\eta$ for some $(n-1)$-form $\eta$.

Now suppose $$\int_M f^\ast\omega=b\int_N\omega$$ for some $b$. You want to show $b=a$. If $c=0$, then $\omega$ is exact, so $\int_N\omega=0$, so we can just set $a=b$. If $c\neq 0$, then $$bc\int_N\omega_0=b\int_N\omega-d\eta=b\int_N\omega=\int_Mf^\ast\omega=\int_Mf^\ast(c\omega_0+d\eta)=c\int_Mf^\ast\omega_0=ca\int_N\omega_0.$$ Since $c\neq 0$ and $\displaystyle\int_N\omega_0\neq 0$, cancelling shows $a=b$.

• Ben West, is $\deg(f)$ still well-defined if the integral is zero for some omegas? I asked about this in a linked question. – Selene Auckland Apr 13 at 14:41

Suppose $M$ is oriented, compact and $\partial N=\emptyset$. All $n$-forms on $N$ are closed, and by Stokes theorem, $\int_N\partial \eta=0$ for any exact form, so that the map $$\int_N:\Omega^n(N)\to\Bbb R$$ automatically factors through the isomorphism $$\int_N:H^n_{dR}(N)\xrightarrow{\sim\,}\mathbb R\,.$$ The conclusion you are after follows from the fact that the linear map $f^*:H^n_{dR}(N)\to H^n_{dR}(M)$ is multiplication by some real number (actually an integer) $a$.

• Olivier Bégassat, is $\deg(f)$ still well-defined if the integral is zero for some omegas? I asked about this in a linked question. – Selene Auckland Apr 13 at 14:42
• @SeleneAuckland it will necessarily be zero for many $n$-forms, all exact forms $\omega=\partial\alpha$ for $\alpha$ an $n-1$ form. You have to look at it as a statement in linear algebra. – Olivier Bégassat Apr 13 at 23:30