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As I understand it, sheaf cohomology is now an indispensable tool in algebraic geometry, but was originally developed to solve problems in algebraic topology. I have two questions about the matter.

Question 1. What is sheaf cohomology? I have a vague idea that it has something to do with right derived functors, but this seems rather far removed from the (admittedly very little) cohomology of (co)chain complexes I do know. I would also like to know why sheaf cohomology appears to be so much more fundamental in algebraic geometry than algebraic topology—for instance, I will be taking second courses in algebraic geometry and topology this coming autumn, but sheaf cohomology only appears in the former, suggesting that perhaps sheaf cohomology is not as relevant in basic algebraic topology. (For example, is there an ‘intuitive’ reason why de Rham cohomology cannot be made to work for algebraic varieties?)

Question 2. Are there any good introductions to sheaf cohomology in a general context? I have tried reading Chapter III of Hartshorne, but very little is getting through, perhaps because I'm not yet comfortable with schemes. A different take—perhaps with an emphasis on manifolds, say—may prove more accessible to me, but since I also need to learn it in the context of algebraic geometry, it would be nice if there were a single text which introduces the theory with applications in both subjects.

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This MO-thread may be of interest. Also, two standard references are Godement and Bredon. The topological context is treated well in Iversen. For a very abstract approach there are the books by Kashiwara and Schapira: Sheaves on manifolds and Categories and Sheaves. –  t.b. Jul 31 '11 at 13:49
(I'm only mentioning the books of KS for the sake of completeness - also, because I know about your interest in category theory in general, and the newer Categories and Sheaves may have many constructions of interest to you. I would definitely not recommend them as an intro to sheaf theory). You are certainly aware of Mac Lane-Moerdijk but IIRC it doesn't have much sheaf cohomology -- if any at all. Also for the algebraic geometry context, you might want to look at Shafarevich. –  t.b. Jul 31 '11 at 13:55
@Theo: Thank you for the suggestions. The absence of sheaf cohomology in Mac Lane and Moerdijk is precisely why I am asking this question—the authors regret its omission yet I do not see its relevance, so I am just very puzzled. I don't read French (yet), unfortunately, and the MR review for Bredon's text seems rather critical. (But perhaps it wasn't a fair review?) [continued] –  Zhen Lin Jul 31 '11 at 14:24
The accepted answer on the MO thread seems to be applicable to cohomology in all its forms—in the sense of measuring ‘obstructions’—and I can imagine that the cohomology of, say, the $\Omega^1$ sheaf might be connected with de Rham cohomology, but how do the right derived functors magically give the same answers, without ‘knowing’ anything about $\Omega^n$ for $n > 1$? –  Zhen Lin Jul 31 '11 at 14:25
@mt_: Here is a much more in-depth review - it looks like it's a nice book. –  Zev Chonoles Jul 31 '11 at 19:49

2 Answers 2

up vote 24 down vote accepted

Sheaf cohomology is the right derived functor of the global section functor, regarded as a left-exact functor from abelian sheaves on a topological space (more generally, on a site) to the category of abelian groups. In fact, one can regard this functor as $\mathcal{F} \mapsto \hom_{\mathrm{sheaves}}(\ast, \mathcal{F})$ where $\ast$ is the constant sheaf with one element (the terminal object in the category of all -- not necessarily abelian -- sheaves, so sheaf cohomology can be recovered from the full category of sheaves, or the "topos:" it is a fairly natural functor.

de Rham cohomology can be made to work for arbitrary algebraic varieties: there is something called algebraic de Rham cohomology (which is the hyper-sheaf cohomology of the analog of the usual de Rham complex with algebraic coefficients) and it is a theorem of Grothendieck that this gives the usual singular cohomology over the complex numbers. Incidentally, sheaf cohomology provides a very simple proof that de Rham cohomology agrees with ordinary cohomology (at least when you agree that ordinary cohomology is cohomology of the constant sheaf, here $\mathbb{R}$) because the de Rham resolution is a soft resolution of the constant sheaf $\mathbb{R}$, and you can thus use it to compute cohomology.

Sheaf cohomology is quite natural if you want to consider questions like the following: say you have a surjection of vector bundles $M_1 \to M_2$: then when does a global section of $M_2$ lift to one of $M_1$? The obstruction is in $H^1$ of the kernel. So, for instance, this means that on an affine, there is no obstruction. On a projective scheme, there is no obstruction after you make a large Serre twist (because it is a theorem that twisting a lot gets rid of cohomology). Sheaf cohomology arises when you want to show that something that can be done locally (i.e., lifting a section under a surjection of sheaves) can be done globally.

$H^1$ is also particularly useful because it classifies torsors over a group: for instance, $H^1$ of a Lie group on a manifold $G$ classifies principal $G$-bundles, $H^1$ of $GL_n$ classifies principal $GL_n$-bundles (which are the same thing as $n$-dimensional vector bundles), etc.

Also, sheaf cohomology does show up in algebraic topology. In fact, the singular cohomology of a space with coefficients in a fixed group is just sheaf cohomology with coefficients in the appropriate constant sheaf (for nice spaces, anyway, say locally contractible ones; this includes the CW complexes algebraic topologists tend to care about). For instance, Poincare duality in algebraic topology can be phrased in terms of sheaves. Recall that this gives an isomorphism $H^p(X; k) \simeq H^{n-p}(X; k)$ for a field $k$ and an oriented $n$-dimensional manifold $X$, say compact. This does not look very sheaf-ish, but in fact, since these cohomologies are really $\mathrm{Ext}$ groups of sheaves (sheaf cohomology is a special case of $\mathrm{Ext}$), so we get a perfect pairing $$ \mathrm{Ext}^p(k, k) \times \mathrm{Ext}^{n-p}(k, k)\to \mathrm{Ext}^n(k,k)$$ where the $\mathrm{Ext}$ groups are in the category of $k$-sheaves. This can be generalized to singular spaces, but to do so requires sheaf cohomology (and derived categories): the reason, I think, that for manifolds those ideas don't enter is that the "dualizing complex" that arises in this theory is very simple for a manifold. You might find useful these notes on Verdier duality, which explains the connection (and which mostly follow the book by Iversen).

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Plus there is Serre's GAGA and/or the Artin comparison theorem (for the etale site) that shows we get the same answer in lots of other situations as well. –  Matt Jul 31 '11 at 20:27
@Zhen: There is a general fact about derived functors: given $F$ (say, right-exact, and on nice abelian categories), if you want to compute the derived functors $R^i F(X)$ for an object $X$, you don't have to use an injective resolution. Rather, if you have a resolution $X \to J^\bullet $ where $J^\bullet $ is a complex of $F$-acyclics, then you can compute $R^i F(X)$ via the cohomology of $F(J^\bullet)$. The point is that the sheaves of $i$forms are acyclic with respect to the global section functor, being soft. This is the reason de Rham cohomology coincides with sheaf cohomology, ... –  Akhil Mathew Aug 1 '11 at 1:53
...even though the de Rham resolution is generally not a resolution by injectives. I would also point out that the cohomology depends very much on the space, so the data in the cohomology comes not from the coefficient group as from the space. For instance, suppose you have a cover $\mathfrak{A}$ of your space such that every intersection is either empty or contractible: then, just by encoding which intersections are empty (data that a constant sheaf can detect!), you can recover the (weak) homotopy type of your space. –  Akhil Mathew Aug 1 '11 at 1:56
The fact I mentioned in my first comment, incidentally, is not that mysterious. Here is an exercise: given a quasi-isomorphism between bounded-below complexes of $F$-acyclics, applying $F$ to the map between both complexes gives a new quasi-isomorphism. (To do this, form the mapping cone, after which you reduce to showing that applying $F$ to a bounded-below, acyclic complex of $F$-acyclics gives an acyclic complex, which is a fun exercise in homological algebra.) Given this, it is easy to check that any $F$-acyclic resolution will do, not necessarily an injective one. –  Akhil Mathew Aug 1 '11 at 1:57
@Zhen: Let's see, this should be in books on homological algebra; for instance, I think it's in the first of Lang's two chapters on the subject (in "Algebra"). This is a general fact about derived functors, but the short explanation is that you can get a quasi-isomorphism of such an acyclic resolution into an injective resolution. So we want to show that applying $F$ preserves the quasi-isomorphism property. To do this, it suffices to show that the mapping cone is acyclic. In other words, the key lemma is that $F$ of a bounded-below, acyclic complex of $F$-acyclics is exact. –  Akhil Mathew Aug 2 '11 at 13:58

Are there any good introductions to sheaf cohomology in a general context?

A source is J.-P. Serre's Faisceaux Algébriques Cohérents. It is very readable. Although the language of algebraic geometry is a bit older, everything carries over to schemes without any problems. A translated version is also available.

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