Connected, oriented and compact smooth manifold can't retract on the boundary

Question

i would like to proof that a compact , connected and oriented $n$-manifold $M$ can't retract onto $\partial M$. My tool is the De Rham cohomology.

En effet, if there exists $f:M\to \partial M$ such that $f\circ \iota_{\partial M}=Id_{\partial M}$, then using the functoriality of cohomological map associated, $\iota^{*}_{\partial M}\circ f^{*}=Id_{H^{p}_{dR}(\partial M)}$, thus for each $p\geq 0$, the map $f^{*}:H^{p}_{dR}(\partial M)\to H^{p}_{dR}(M)$ is injective. So, i imagine there is a $p$ where the map injection fails, but with the hypotheses, with $p=n$, $f^{*}\equiv 0$, and i stopped here. Any tips?

Answer 1

First of all if you want to prove your states with the de Rham cohomology you get only that does not exist a smooth retract of $M$ to his boundary because you can use the de Rham cohomology only if your map is smooth.

If $H^{n-1}_{dr}(M)=0$ than the identity map for $p=n-1$ would be zero because $f^*$ is zero but is a contradiction because $\partial M$ is a compact without boundary (n-1)-manifolds and so by Poincarè duality you have that:

$H^{n-1}_{dr}(\partial M)=\mathbb{R}$

Answer 2

oriented compact manifolds have trivial $H^{n-1}$, but this is not so for $\partial M$, which has fundamental class $\mathbb R$. A retract should be a map $r:M \to \partial M$ so that $r \circ i=id$, where $id$ is the identity map.

Hence by functoriality on $H^{n-1}$ we have that $i^* \circ r^*=i^*(0)$ which can not be $id^*:\mathbb R\to \mathbb R$.