Functoriality of Poincaré duality and long exact sequences - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:40:35Z http://mathoverflow.net/feeds/question/25593 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25593/functoriality-of-poincare-duality-and-long-exact-sequences Functoriality of Poincaré duality and long exact sequences Saul Glasman 2010-05-22T18:00:07Z 2010-05-22T19:08:51Z <p>Hi all,</p> <p>Today I was playing with the cohomology of manifolds and I noticed something intriguing. Although I might just have been caught out by a couple of enticing coincidences, it feels enough like there might be something going on that I thought I'd put it out here. </p> <p>Let $M$ be an $n$-manifold with boundary $\partial M$. We write out the long exact homology sequence for the pair $(M, \partial M)$:</p> <p>$$\cdots \to H_k(M) \to H_k(M, \partial M) \to H_{k - 1}(\partial M) \to \cdots$$</p> <p>Let's apply Poincaré duality termwise, and keep the arrows where they were out of sheer faith. What we get is</p> <p>$$\cdots \to H^{n - k}(M, \partial M) \to H^{n - k}(M) \to H^{n - k}(\partial M) \to \cdots$$</p> <p>Surprisingly, this is the long exact <i>cohomology</i> sequence for the pair $(M, \partial M)$! To my mind, two things here are weird. The first is that intuitively, any functoriality properties Poincaré duality possesses should be arrow-reversing. The second is that we have a shift - but not a shift by a multiple of 3. So the boundary map in the homology sequence 'maps' to something that doesn't change degree in the cohomology sequence. </p> <p>Let's play the same game with the Mayer-Vietoris sequence. For simplicity, suppose now $M$ is without boundary. Write $M = A \cup B$ where $A$ and $B$ are $n$-submanifolds-with-boundary of $M$ and $A \cap B$ is a submanifold of $M$ with boundary $\partial A \cup \partial B$. Then we have</p> <p>$$\cdots \to H_k(A \cap B) \to H_k(A) \oplus H_k(B) \to H_k(M) \cdots$$</p> <p>Hitting it termwise with Poincaré duality, and cruelly and unnaturally keeping the arrows where they are once again, we get</p> <p>$$\cdots \to H^{n - k}(A \cap B, \partial A \cup \partial B) \to H^{n - k}(A, \partial A) \oplus H^{n - k}(B, \partial B) \to H^{n - k}(M) \to \cdots$$</p> <p>This looks unfamiliar, but by looking at cochains it's not hard to see that there actually is a long exact sequence with these terms. However, this time we don't have the weird shift.</p> <p>Now is there anything going on here, or just happenstance? Is there really a sense in which Poincaré duality is functorial with respect to long exact sequences? If so, what's the 'Poincaré dual' of the long exact sequence of the pair $(M, N)$ where $N$ is a tamely embedded submanifold of $M$? </p> <p>Edit: Realised that in the final l.e.s. the arrows should actually go the other way, which is slightly less impressive. Even so...</p> http://mathoverflow.net/questions/25593/functoriality-of-poincare-duality-and-long-exact-sequences/25598#25598 Answer by Greg Kuperberg for Functoriality of Poincaré duality and long exact sequences Greg Kuperberg 2010-05-22T19:08:51Z 2010-05-22T19:08:51Z <p>One of the standard proofs of Poincaré duality, at least for those manifolds that have handle decompositions, provides a reason for some of these naturality properties. Every piecewise linear manifold, or every smooth manifold, has a handle decomposition, and many but not all topological manifolds also do. (Amazingly enough, the only exceptions are in 4 dimensions.) A handle decomposition gives rise to two different CW cellulations on the manifold, one using cores and the other using co-cores. Then this proof of Poincaré duality posits that the CW chain complex of one cellulation is identical to the CW cochain complex of the other cellulation.</p> <p>You can extend this coincidence of chain complexes to both of your examples, the Mayer-Vietoris sequence and the exact sequence of a pair. Obtaining identical chain complexes also gives you other information, for instance that the Bockstein maps are the same.</p>