If we have a compact orientable manifold $M$, we know that $\partial M$ is not a deformation retract of $M$. This follows from Poincaré Duality or Stokes Theorem.
If we take away compactness, this is not true. For instance, take the half upper plane.
I`m interested in an example of a compact one, but not orientable, such that $\partial M$ is a deformation retract of $M$. Is this possible? If so, what is an example?
Best Answer
For any non-closed $n$-manifold $M$, $H_n(M;G) = 0$ for all groups $G$. See Hatcher theorem 3.29. (To pass from a compact manifold with boundary to a noncompact one, glue on a collar neighborhood of the boundary.) I believe that morally this is because compact manifold $M$ with boundary is homotopy equivalent to an $(n-1)$-dimensional CW complex; I'm quite certain this is true but don't remember a reference.
Now for a compact $n$-manifold with boundary $M$, regardless of orientation woes, $H_n(M,\partial M;\mathbb Z/2) \cong \mathbb Z/2$. You can either prove this by hand or use Lefschetz duality.
Now the bit of the long exact sequence of the pair $(M,\partial M)$ we care about is $$0 \to H_n(M,\partial M) \to H_{n-1}(\partial M) \to H_{n-1}(M).$$ If there was a retract $M \to \partial M$, then the map $H_{n-1}(\partial M) \to H_{n-1}(M)$ would be injective. But the kernel of this map is the image of the injective map $\mathbb Z/2 \cong H_n(M,\partial M) \to H_{n-1}(\partial M)$. There's your contradiction.
If you want to avoid Poincare duality, you can get a weaker result ($M$ doesn't deformation retract onto the boundary; you can't say anything about retracting) more easily using the link and outline Jim Belk posted in the comments above. For pass to the double $DM$; if $M$ deformation retracted onto $\partial M$, then $DM$ deformation retracts onto $\partial M$; the retraction is not surjective hence has mod 2 degree zero; but the mod 2 degree of the identity map is one, and mod 2 degree is a homotopy invariant.