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That there is no $C^1$ retraction from a compact, oriented manifold to its boundary is a common lemma in proving a weaker version of the Brouwer fixed point theorem. I recall seeing in class a simple proof of this using Stokes' theorem, but I'm unable to find a satisfactory proof online. From what I've found, the proofs start with a oriented k-manifold $M$ and a $k-1$-form $\omega$ such that $\int_{\partial M} \omega > 0.$ Then we assume we have a $C^1$ retraction $f: M \to \partial M$, and Stokes' theorem can be used to show that $$\int_{\partial M} \omega = \int_M f^*(d \omega) = 0$$ But I don't understand why the last equality holds, or if there is an error in what I'm finding.

Thank you in advance.

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Since $f$ is a retraction, $f^*\omega\in\Omega^{n-1}(M)$ has the property that $f^*\omega|_{\partial M}=\omega$. Since $M$ is compact and oriented, you can apply Stoke's theorem to get $\int_{\partial M}\omega=\int_{\partial M}f^*\omega=\int_Md(f^*\omega)$. But $d(f^*\omega)=f^*(d\omega)$ and $d\omega=0$ since $\dim(\partial M)=n-1$.

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  • $\begingroup$ I'm so sorry, but I'm not sure if I understand why $dim(\partial M) = n-1$ implies $d\omega = 0$. $\endgroup$ – florence Jun 23 '16 at 7:15
  • $\begingroup$ Since $\omega$ is an $n-1$-form on $\partial M$, $d\omega$ is an $n$-form on $\partial M$ and above the dimension, all forms are zero. $\endgroup$ – Andreas Cap Jun 23 '16 at 7:17
  • $\begingroup$ Thank you! This makes sense now. $\endgroup$ – florence Jun 23 '16 at 7:29

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