Poincaré duality says that for a compact, orientable manifold without boundary the $k$th and $(n-k)$th homology groups are isomorphic.

For domains with boundary, it's easy to construct examples where this statement is no longer true. For instance, consider a topological disk constructed as a CW complex with one cell in each dimension, i.e., take a 0-cell $x$, glue both endpoints of a 1-cell $\gamma$ to $x$, and finally glue in a 2-cell $A$ homeomorphically. From here it's easy to check that duality fails:

$$H_2 = \mathrm{ker}\ \partial_2/\mathrm{im}\ \partial_3 = \mathrm{ker}\ \partial_2 = \emptyset,$$

since every 2-cell (namely: $A$) has boundary; but we also have

$$ H^0 = \mathrm{ker}\ \delta_0/\mathrm{im}\ \delta_{-1} = \mathrm{ker}\ \delta_0 = nx,\ n \in \mathbb{Z}$$

since the 0-cell $x$ has empty coboundary, i.e., there is no 1-cell with boundary $x$ (consider that $\partial_1 \gamma = x - x = 0$). Hence, $H_2 \ne H^0$.

However, this example does not provide much insight -- we see that duality fails, but why? In other words, what is the essential difference between what is captured by homology and cohomology when the domain has boundary? Do homology and cohomology together capture everything about the topology of a manifold with boundary? What information can we read off from the homology versus the cohomology independently? (Maybe there's a nice categorical way of thinking about these questions?)


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    $\begingroup$ There's a version of Poincare duality for manifolds with boundary; check Hatcher for instance. Note that (co)homology doesn't capture everything about the topology of a manifold with or without boundary, for instance the "Poincare homology sphere" is a counterexample. $\endgroup$ – Aaron Mazel-Gee Jun 4 '12 at 17:47
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    $\begingroup$ You're misquoting Poincaré duality again (compact, oriented). I also feel like you're being surprised by the wrong thing. The question, to me, isn't "what about manifolds with boundary makes the homology and cohomology not Poincaré dual?" but "what about compact oriented manifolds without boundary makes the homology and cohomology Poincaré dual?!" That is the part that seems to me surprising and in need of explanation. $\endgroup$ – Qiaochu Yuan Jun 4 '12 at 17:55
  • $\begingroup$ The version of Poincaré duality for manifolds with boundary referenced above is called Poincaré-Lefschetz duality or sometimes just Lefschetz duality. $\endgroup$ – Henry T. Horton Jun 4 '12 at 17:55
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    $\begingroup$ In particular, Poincaré duality implies that you can read off the dimension of a compact oriented manifold without boundary from its homotopy type, which is surprising as this fails badly if either the compactness or boundarylessness hypotheses are dropped (e.g. given any manifold $M$ the manifolds $M, M \times \mathbb{R}$ and $M \times [0, 1]$ have the same homotopy type but the latter two are one dimension higher and the second is compact if $M$ is but has boundary). $\endgroup$ – Qiaochu Yuan Jun 4 '12 at 18:00
  • $\begingroup$ You're right; I should not have omitted "compact and orientable" -- now fixed. $\endgroup$ – fuzzytron Jun 4 '12 at 22:32

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