This is exercise 3.3.33 in Hatcher.

Show that if $M$ is a compact contractible $n$-manifold then $\partial M$ is a homology $(n-1)$-sphere; that is, $H_i(\partial M; \mathbb{Z}) \approx H_{i}(S^{n-1}; \mathbb{Z})$ for all $i$.

I have a proof for this in the case that $M$ is orientable using Lefschetz duality. I don't know how to prove it in the case that $M$ is not orientable.

Is there some reason that a compact contractible manifold (with boundary) needs to be orientable?

If not, how should I go about proving this without orientability?

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    $\begingroup$ Hint: Contractibility says something about $\pi_1$. Non-orientability says something different about $\pi_1$. $\endgroup$ – Jason DeVito Oct 6 '18 at 1:13
  • $\begingroup$ @JasonDeVito Jason, I thought about that earlier. I know the theorem for manifolds without boundary but does it for sure hold with boundary? $\endgroup$ – TuoTuo Oct 6 '18 at 1:44
  • $\begingroup$ The theorem is for any topological space where the "usual" covering space theory applies. So, semi-locally simply connected, locally path connected, etc. Whatever those usual hypotheses are. Certainly all topological manifolds have them, as do all CW complexes, etc. (maybe I need to assert finitely many cells in each dimension? probably not - but I don't usually think of things nastier than that) $\endgroup$ – Jason DeVito Oct 6 '18 at 1:49
  • $\begingroup$ @JasonDeVito I don't really know what it means for a space that is not some sort of manifold to be orientable, but it sounds right that a manifold with boundary should be orientable if it's simply connected. $\endgroup$ – TuoTuo Oct 6 '18 at 1:54
  • $\begingroup$ Yeah, good point about the orientability requiring a nice space. In fact, now that I think about it more, I'm not positive about the connection between $\pi_1$ and orientability when there is a boundary involved.... $\endgroup$ – Jason DeVito Oct 6 '18 at 2:09

A manifold with boundary is defined to be orientable if $int(M) = M \setminus \partial M$ is orientable. Since simply connected manifolds are orientable, it suffices to show that $int(M)$ is simply connected.

In fact $int(M)$ is contractible. It is known that $\partial M$ has an open collar neighborhood $N$ in $M$ (see Hatcher, Proposition 3.42). Let $h : \partial M \times [0,1) \to N$ be a homeomorphism such that $h(x,0) = x$ for all $x$. Define $M' = M \setminus h(\partial M \times [0,1/2)) \subset int(M)$. This is a homeomorphic copy of $M$. But obviously $M'$ is a strong deformation retract of $int(M)$, hence $int(M) \simeq M' \simeq \ast$.

By the way, open collar neighborhoods exist for any manifold with boundary. Hatcher proves it only in the compact case.

  • $\begingroup$ I would define the orientation on a manifold with boundary to be an orientation on the tangent space at every point, with a compatibility condition either using interior or boundary charts. But it is a quick exercise to see that our definitions are equivalent. $\endgroup$ – user98602 Oct 6 '18 at 13:23
  • $\begingroup$ @MikeMiller You are right, that is the better approach, at least for differentiable manifolds. For topological manifolds without boundary one can use the tangent microbundle instead. For topological manifolds with boundary this approach requires to consider collars, and an orientation of $M$ in the micro-bundle sense is essentially the same as an orientation of $int(M)$. Anyway, Hatcher avoids the use of bundles and defines an orientation of $M$ to be one of $int(M)$ (see p. 253 of his book).. $\endgroup$ – Paul Frost Oct 6 '18 at 15:22
  • $\begingroup$ @MikeMiller for that case isnt it enough to say that tangent bundle is trivial for contractible spaces (over a point)? $\endgroup$ – Andres Mejia Oct 7 '18 at 0:03
  • $\begingroup$ @PaulFrost Interesting, good point. I think one could hack together some notion of local homology at a point $x \in A$ of a pair $(X,A)$ and use this to define boundary orientation more like the interior case. But of course this is wholly unnecessary. $\endgroup$ – user98602 Oct 7 '18 at 19:22

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