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Background

Exercise 2.1.16b in Hartshorne (homework!) asks you to prove that if $0 \rightarrow F \rightarrow G \rightarrow H \rightarrow 0$ is an exact sequence of sheaves, and F is flasque, then $0 \rightarrow F(U) \rightarrow G(U) \rightarrow H(U) \rightarrow 0$ is exact for any open set $U$. My solution to this involved the axiom of choice in (what seems to be) an essential way.

Essentially, you are asking to $G(U) \rightarrow H(U)$ to be surjective when you only know that $G \rightarrow H$ is locally surjective. Ordinarily, you might not be able to glue the local preimages of sections in $H(U)$ together into a section of $G(U)$, but since $F$ is flasque, you can extend the difference on overlaps to a global section. This observation deals with gluing finitely many local preimages together. Zorn's lemma enters in to show that you can actually glue things together even if the open cover of $U$ is infinite.

Now, I have not really studied sheaf cohomology, but the idea I have is that it detects the failure of the global sections functor to be right exact. So if you can't even show sheaf cohomology vanishes for flasque sheaves without the axiom of choice, it seems like a lot of the machinery of cohomology would go out the window.

Now, just on the set theoretic level, it seems like there is something interesting going on here. Essentially the axiom of choice is a local-global statement (although I had never thought of it this way before this problem), namely that if $f:X \rightarrow Y$ is a surjection you can find a way to glue the preimages $f^{-1}({y})$ of a surjection together to form a section of the map $f$.

This brings me to my

Questions

Can the above mentioned exercise in Hartshorne be proven without the axiom of choice?

How much homological machinery depends on choice?

Have any reverse mathematicians taken a look at sheaf cohomology as a subject to be "deconstructed"?

Have any constructive set theorists thought about using cohomological technology to talk about the extent to which choice fails in their brand of intuitionistic set theory? (it seems like topos models of such set theories might make the connection to sheaves and their cohomology very strong!)

My google-fu is quite weak, but searches for "reverse mathematics cohomology" didn't seem to bring anything up.

flag	asked Jan 17 at 17:40 Steven Gubkin 1,229●3●16

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Why in the world does writing the text "make a connection" automatically create a link to an Oprah book? – Steven Gubkin Jan 18 at 4:06

François G. Dorais · Accepted Answer · 2010-01-17 18:32:03Z UTC

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I don't have Hartshorne, so I can't address the specifics of this case. However, there is a very interesting paper by Andreas Blass Cohomology detects failures of the Axiom of Choice (TAMS 279, 1983, 257-269), which addresses questions of this type and should at least put you on the right track.

answered Jan 17 at 18:32

François G. Dorais
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	That paper looks like exactly what I am after! I will try not to get too excited about it until I have finished writing up my part of our paper though. – Steven Gubkin Jan 17 at 18:37
	Actually, I don't think it has exactly what you want, but it will at least be inspirational... – François G. Dorais Jan 17 at 18:44
	I have obviously only glanced it over, but it seems to address thelast point to some extent. I think there is a lot that could be said here, and that paper is a good starting point. – Steven Gubkin Jan 17 at 18:50

VA · Answer 2 · 2010-01-17 18:33:27Z UTC

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A lot more of algebraic geometry will go out the window without the axiom of choice, not just sheaf cohomology. To begin with, proving that every ring has a prime ideal requires Zorn's lemma. So you couldn't even prove that $\operatorname{Spec} R$ is nonempty without it. So you don't have spaces to work with in the first place.

answered Jan 17 at 18:33

VA
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I think the functor of points approach still works without ever looking at spectra, but I am not sure how far you can go there without mentioning prime ideals. – Steven Gubkin Jan 17 at 18:39

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The existence of prime ideals is often weaker than the Axiom of Choice. In any case, a lot can be done with pointfree spaces. See Johnstone's Stone Spaces, for example. – François G. Dorais Jan 17 at 18:42

@ F.G.Dorais: what do you mean? Existence of primes for some special classes of rings? Which ones? Where is that explained? – VA Jan 18 at 0:39

@VA: For commutative rings, the existence of prime ideals is a simple consequence of the Compactness Theorem for propositional logic. In turn, the Compactness Theorem is equivalent to the existence of prime ideals for Boolean algebras. The fact that this is weaker than AC was shown by Halpern and Levy in the 1960s, see Tom Jech's The Axiom of Choice for details. – François G. Dorais Jan 18 at 1:17

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In the absence of choice, one should really be using locales instead of topological spaces anyway. (For instance, Tychonoff's theorem for topological spaces is equivalent to choice, whereas Tychonoff's theorem for locales is just true, even in intuitionistic logic without any choice whatsoever.) The "prime spectrum" of a ring is always a perfectly well-behaved locale, although without choice it may not be a space. (Algebraic geometers may be more comfortable talking about sites or topoi rather than locales.) – Mike Shulman Jan 18 at 2:38

Ben Webster · Answer 3 · 2010-01-17 18:08:09Z UTC

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On any affine scheme and thus on any scheme with a finite affine cover (which I think most people would find a reasonable class of schemes to restrict to), any open cover has a finite subcover. This property is usually called quasi-compact rather than compact for technical reasons.

I don't have Hartshorne with me, so I can't recite chapter and verse, but I know this is discussed somewhere in Hartshorne; it's also covered by this nLab entry, though I suspect that's more technical than you're looking for.

answered Jan 17 at 18:08

Ben Webster♦
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	I believe that those "technical reasons" are that some people (most notably Bourbaki) require compact spaces to be Hausdorff. – Andy Putman Jan 17 at 18:11
	I believe it also has to do with the fact that "quasi-compact" is a property of a morphism of schemes, not just of topological spaces. – Ben Webster♦ Jan 17 at 18:12
	The question about sheaves is about sheaves of general topological space. Actually, by what you have said, the case you mentioned is not interesting as far as my questions go. – Steven Gubkin Jan 17 at 18:14
	A side comment: As Andy wrote, for Bourbaki, "compact" means "compact Hausdorff", while "quasi-compact" means "compact, but not necessarily Hausdorff". There are certain absolute properties of schemes that can also be applied to morphisms ("quasi-compact, affine, ... "), but this has nothing to do with the "quasi" in "quasi-compact", as far as I know. – Emerton Jan 18 at 5:08

Reverse mathematics of (co)homology?

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