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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?

Aloizio Macedo
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    How about a Mobius strip, modulo concerns about what the definition of "is" is? – Neal Jul 13 '15 at 15:59
  • When $M$ is the Möbius strip, the space $\partial M = S^1$ is a deformation retract of the space $M$ (indeed, the Möbius band deformation retracts onto its core, which is a circle, just like the boundary of the Möbius band). But the Möbius band doesn't deformation retract onto its boundary (for degree reasons essentially). I believe this is what @Neal meant? In which case you need to clarify your question to say what scenario you're looking for. This example also shows that you're not necessarily going to be able to disprove it just by general considerations (like Poincaré duality). – Najib Idrissi Jul 13 '15 at 16:16
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    Use the long exact sequence of the pair $(M,\partial M)$ with $\mathbb Z/2$ coefficients. All you need to know to finish this is that $H_n(M,\partial M;\mathbb Z/2) \cong \mathbb Z/2$ and $H_n(M) = 0$ (both of which are true for any compact manifold with boundary, orientable or not). This proves that in the non-oriented case you still can't even retract onto the boundary, much less deformation retract. –  Jul 13 '15 at 16:55
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    Here's another proof: if $M$ deformation retracts onto $\partial M$, then so does the double of $M$. But by this question, no compact manifold without boundary can deformation retract onto a proper subset of itself. – Jim Belk Jul 13 '15 at 17:41
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    Related: Is there a retraction of a non-orientable manifold to its boundary?, Retraction to the Boundary on Compact Manifold, M is a compact manifold with boundary N,then M can't retract onto N.. – Najib Idrissi May 25 '16 at 08:51

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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.