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Jan 20, 2016 at 22:28 comment added Fernando Muro Well, the tautological iff looks sufficient to me, and it is induced by a chain map since, over a PID, any chain complex is quasi-isomorphic to its homology.
Jan 20, 2016 at 19:55 comment added Najib Idrissi Well it feels rather weird to say that something like $0 \to \mathbb{Z} \xrightarrow{=} \mathbb{Z} \to 0 \to 0$ satisfies Poincaré duality...
Jan 20, 2016 at 18:13 comment added Freddy Yes, but I don't want to specify a map.
Jan 20, 2016 at 17:46 comment added Qiaochu Yuan That seems like a weak condition. Wouldn't you rather these isomorphisms be induced by a map from the chain complex to the dual chain complex?
Jan 20, 2016 at 17:37 review First posts
Jan 20, 2016 at 17:45
Jan 20, 2016 at 17:35 history asked Freddy CC BY-SA 3.0