Singular Homology/Cohomology as a derived functor? - MathOverflow

Singular Homology/Cohomology as a derived functor?

2011-05-29T22:32:52Z

Hello, Learning some Alg.geometry and Sheaf theory, I got used to the notion that cohomology arises naturally as a derived functor of some sort.

This has led me thinking, singular cohomology, from algebraic topology, was never defined (In all books i've checked) as a derived functor, but just by giving cycles and boundaries. I could not figure out by myself any reasonable functor whose derived functors yield singular cohomology, So I pose this question out here.

I hope this might shed some more insight on what singular cohomology actually measures.

Thanks

Answer by David Roberts for Singular Homology/Cohomology as a derived functor?

2011-05-30T00:27:37Z

Just some high-level thoughts...

Singular cohomology sits more naturally in the setting of (Quillen) model categories, using the chain of Quillen functors $Top \to sSet \to sMod_R \to Chain_R$. Quillen invented/discovered model categories in order to describe 'non-abelian derived functors', that is homotopical algebra rather than homological algebra. Essentially this means functors on the homotopy category, but described using objects of the original category (think: homotopy type represented by spaces). Reciprocally, we now can think about derived functors as being homotopical in nature.

If I was Urs Schreiber* I would now say that really this was all about $(\infty,1)$-categories and functors between them, but that is probably beyond the scope of the question. Or perhaps not! If so, ask away.

*A good colleague of mine :)

Answer by Greg Friedman for Singular Homology/Cohomology as a derived functor?

2011-05-30T01:51:57Z

Elaborating a bit on Qiaochu Yuan's comment, if X is "nice enough" and $\mathcal{R}$ is the constant sheaf with stalks in $R$, then the singular cohomology agrees with the derived functor of the global section functor: $H^\ast(X;\mathcal{R})\cong H^\ast(X;R)$. This result is scattered throughout Bredon's book on sheaf theory, though I grant that it's not super-intuitive there. Alternatively, it's not hard to show the sheafification of the presheaf complex of singular cochains $U\to C^\ast(U;R)$ is a resolution of $\mathcal{R}$ and it can be shown to be homotopically fine, by a dual argument to the proof that the sheaf complex of singular chains is homotopically fine. Furthermore the presheaf of cochains is conjunctive and, while it's not a mono-presheaf, the cohomology with zero supports of the cochain presheaf is trival. Putting all those things together, the singular cohomology is isomorphic to the hypercohomology of the cochain sheaf complex, which is a derived functor of the global section functor, though in the "hyper" sense. Similarly things can be done with homology, the sheaf of germs of singular chains being homotopically fine. Swan's book on sheaf theory is a good reference for that.

Answer by Tim Porter for Singular Homology/Cohomology as a derived functor?

2011-05-30T12:00:57Z

On seeing the question, I knew that I had seen something like this. Rinehart had a TAMS paper: 1972 called SINGULAR HOMOLOGY AS A DERIVED FUNCTOR. His interpretation of derived functor is, of course, that of his era but it does give a setting that may be adapted to a more modern treatment. Following up on David's reply, it is also possible to use Quillen's treatment of cohomology from the original Homotopical Algebra SL notes, together with ideas on monadic cohomology (the Triples volume of SLN paper by Applegate and Tierney, plus ideas from Mike Barr) to get another take on the question.... but without the various volumes in front of me I will not attempt to do that here and now. :-)

Answer by Saul Glasman for Singular Homology/Cohomology as a derived functor?

2011-05-30T15:49:49Z

Here's an elaboration of David Roberts' answer, a perspective that can be found in Quillen's 'Homotopical Algebra'. Quillen thought of homology as "derived abelianisation", and the abelianisation of a space (by which I mean simplicial set) can only be the free simplicial abelian group on that space, seeing as it's left adjoint to the forgetful functor $sAb \to sSet$.

Composing the free simplicial abelian group functor with the Dold-Kan correspondence - an equivalence of categories between simplicial abelian groups and chain complexes - gives a functor $sSet \to Ch$, which happens to coincide (more or less) with the singular complex functor. If we endow $Ch$ with the injective model structure this is a left Quillen functor. By Quillen's philosophy, homology of spaces should be the total left derived functor of this functor. But the word 'derived' is sort of vacuous here, because all simplicial sets are cofibrant, and this negates the need to explicitly use any model category language. This is why we think of the singular complex as the "homology object" of a space.