I attempted a problem from An introduction to differential manifolds by Jacques Lafontaine, but was wondering if I had made a mistake anywhere. It is question 20 from Chapter 7, pg. 321.
Let $M$ be a compact, orientable, $(n+1)$-dimensional manifold with connected boundary $\partial M$, and let $f$ be a smooth map from $M$ to a compact, orientable manifold of dimension $n$. Show that $\deg(f|_{\partial M})=0$.
My attempt: Let $f:M\to N$ be smooth where $M$ is as above and $N$ is a compact, orientable $n$-dimensional manifold. Since $M$ is $(n+1)$-dimensional, $\partial M$ is a compact, orientable $n$-dimensional manifold. Let $\omega\in \mathcal{A}^n(N)$ be an orientation form on $N$. Then $f|_{\partial M}^*\omega \in \mathcal{A}^n(\partial M)$ and, in particular, $d(f|_{\partial M}^*\omega)=0$ since $\mathcal{A}^{n+1}(\partial M)=\{0\}$. Hence
$$\deg (f|_{\partial M})\int_N\omega = \int_{\partial M}f|_{\partial M}^*\omega = \int_M d(f|_{\partial M}^*\omega) = 0$$
where the second equality is by Stoke's Theorem. However, $\int_N\omega\not=0$ since $\omega$ is an orientation form on $N$, so we have that $\deg(f|_{\partial M})=0$ as desired.
My apologies if I am making any major misconceptions or wrongfully over-simplifying the question. My main concern is that I am missusing Stoke's Theorem. Any help is greatly appreciated!
Best Answer
This proof is not quite right. You need a form defined on all of $M$ in order to apply Stokes's Theorem. By your logic, any time you integrate a form on the boundary you would get $0$. So what is the easy fix here? How do you get an $n$-form on all of $M$?