Does exists a proof of the Poincaré Duality version for non-compact manifolds without using the Zorn's Lemma? I know that there is a proof using the Whitney embedding theorem, but I don't know this theorem's proof, so I don't know if it uses the Zorn's Lemma too.
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3$\begingroup$ Where is Zorn's lemma used in the proof you have in mind? $\endgroup$– Qiaochu YuanCommented May 13, 2016 at 7:20
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$\begingroup$ Sorry, mistake, editing. $\endgroup$– iam_agfCommented May 13, 2016 at 7:25
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$\begingroup$ I meant the Poincaré Duality for non-compact manifolds. The proof I read is the one from Hatcher. $\endgroup$– iam_agfCommented May 13, 2016 at 7:27
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3$\begingroup$ Yes, Poincare's original proof works just fine -- for manifolds with triangulations. Technically, Whitehead's triangulations proof uses Whitney's theorem but triangulability is quite different in character. Whitney's theorem has fairly direct proof. I encourage you (really, everyone) to learn the proof. It is analogous to quite a few other basic theorems. I suppose the handlebody proof of Poincare duality does not use Zorn's lemma in any heavy-handed way. $\endgroup$– Ryan BudneyCommented May 13, 2016 at 7:34
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$\begingroup$ Can you add the proof you're telling me to read it please? $\endgroup$– iam_agfCommented May 13, 2016 at 7:37
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2 Answers
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Often the definition of a manifold includes a condition to exclude strange manifolds that are "very large" such as the Prüfer surface. One condition commonly assumed is that the topology on the manifold is second countable. This is necessary for embedding the manifold in Euclidean space for example. Second countability implies that there is a countable cover by coordinate charts, and in this case the proof of Poincaré duality in my book works without Zorn's Lemma, as stated at the end of step (3) of the proof on page 248. Perhaps I should have assumed second countability and skipped the part of the proof using Zorn's Lemma, but I thought it was kind of interesting that duality holds in the more general setting if one is willing to use Zorn's Lemma. I don't know whether the general case can be proved without Zorn's Lemma (or something equivalent). As a test case one could look for a Zornless proof for the Prüfer surface. Even simpler, one could look at the disjoint union of an uncountable number of copies of Euclidean space.
answered May 13, 2016 at 15:29
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$\begingroup$ I can only say that I feel honored to be answered by you, Dr. Hatcher. I will take a look about that proof for the Prüfer surface you're telling. $\endgroup$– iam_agfCommented May 13, 2016 at 16:55
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A slight modification of Hatchers Proof gives you what you want: Hatcher relies on the fact that if $M$ is a union of opens totally ordered by inclusion, which satisfy PD, then $M$ does, too. It is easy to see that you can weaken that to the opens only being directed. Also, another case tells you that all finite intersections of coordinate balls satisfy PD, so all finite unions do too. Then, at the end instead of using Zorn's Lemma you only have to look at the cover of all finite unions of coordinate balls, which are obviously directed w.r.t. inclusion.
