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The question title says it all. For the record, I have no reason to believe that this is true, but my question has a bit of a background.

I am reading Ramanan's Global Calculus book because I am interested in the isomorphism between singular cohomology with coefficients in a ring $R$ and sheaf cohomology with respect to the constant sheaf $\underline R$ (page 114, Theorem 4.14). He only assumes his topological space $X$ is locally contractible, but he quotes a result (page 8,Lemma 1.14) which assumes his space is hereditarily paracompact ; the result says the sheafification map of a presheaf $\mathcal F(U) \to \mathcal F^+(U)$ is surjective for all open subsets $U \subseteq X$, assuming the existence part of the glueing axiom.

So I was wondering if he just wrote somewhere he would make that assumption for a long time, forgot to assume it, or if the question in my title has a positive answer, which to be honest, would surprise me! But hey, until I have a counter-example... who knows.

Eric Wofsey
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Patrick Da Silva
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  • I just asked a similar question for exactly the same reason! (Also because I am dissatisfied with some other proofs of the equivalence of singular/sheaf cohomology, which seem to rely on the surjectivity of the sheafification map, though they assume only local contractibility.) http://math.stackexchange.com/questions/1794725/detail-in-the-proof-that-sheaf-cohomology-singular-cohomology?noredirect=1#comment3664415_1794725 – Eric Auld May 22 '16 at 04:16
  • Also, I think I found a serious mistake in that chapter http://math.stackexchange.com/questions/1795745/barycentric-subdivision-proof – Eric Auld May 22 '16 at 23:09

3 Answers3

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A locally contractible space need not even be paracompact. For instance, the long line is locally homeomorphic to $\mathbb{R}$, but not paracompact.

Eric Wofsey
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  • Same question as the one I asked Captain Lama! – Patrick Da Silva Apr 21 '16 at 17:56
  • The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $\mathcal{F}(X)\to\mathcal{F}^+(X)=\mathcal{F}({0})\times\mathcal{F}({1})$ to be surjective. – Eric Wofsey Apr 21 '16 at 17:59
  • @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now. – Patrick Da Silva Apr 21 '16 at 18:04
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The long line should be a counter-example : it is known that it's not paracompact (see https://en.wikipedia.org/wiki/Long_line_%28topology%29), but it's a manifold if we omit the "second countable" condition from the definition, so in particular it's locally homeomorphic to an open interval, and thus locally contractible.

Captain Lama
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  • Do you have an idea about what happened in the book? Thanks for bringing up the example though! – Patrick Da Silva Apr 21 '16 at 17:55
  • @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified. – Henno Brandsma Apr 22 '16 at 07:53
  • @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain. – Patrick Da Silva Apr 22 '16 at 12:04
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To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility, in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3), where he also explains how Ramanan uses the hereditary paracompactness condition.

Dmitri P.
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