I need help to understand Shifrin's argument that he uses to prove the hairy ball theorem from Stokes's theorem. This is on on pages 403 and 404 in his Multivariable Mathematics book and the whole thing is quoted verbatim below.
[...] Let $S^n$ denote the n-dimensional unit sphere, $S^n=\{\textbf{x}\in \mathbb{R}^{n+1} : ||x||= 1\}$, and $D^n$ the closed unit ball, $D^n=\{\textbf{x}\in \mathbb{R}^n : ||x||\le 1\}$. (Then $\partial D^{n+1}=S^n$)
$\textbf{Lemma 7.1}$ There is an n-form $\omega$ on $S^n$ whose integral is non-zero.
$\textbf{Proof}$ It is easy to check that the volume form $$\omega = \sum_{i=1}^{n+1}(-1)^{i-1}x_idx_1\wedge...\wedge\widehat{dx_i}\wedge...\wedge dx_{n+1}$$ is such form.$\blacksquare$
$\textbf{Theorem 7.2}$ There is no smooth function $\textbf{r}: D^{n+1} \rightarrow S^n$ with the property that $\textbf{r(x)=x}$ for all $\textbf{x} \in S^n$
$\textbf{Proof}$ Suppose there were such an $\textbf{r}$. Letting $\omega$ be an $n$-form on $S^n$, as in Lemma 7.1, we have $$\int_{S^n}\omega=\int_{S^n}\textbf{r}^*\omega=\int_{D^{n+1}}d(\textbf{r}^*\omega)=\int_{D^{n+1}}\textbf{r}^*(d\omega)=0$$ inasmuch as the only $(n+1)$-form on an an n-dimensional manifold is $0$ (and hence $d\omega=0$. But this is a contradiction since we chose $\omega$ with a nonzero integral.$\blacksquare$
Now $d\omega$ is an $n+1$-form and $D^{n+1}$ is an $n+1$-dimensional manifold, so what is the $n$-dimensional manifold over which $d\omega$ integrates to $0$ and Shifrin is talking about? Thank you.