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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

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    $\begingroup$ For nice spaces singular cohomology agrees with certain sheaf cohomology groups, which come from the global sections functor. See en.wikipedia.org/wiki/… . $\endgroup$ Commented May 29, 2011 at 22:46
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    $\begingroup$ The difference between the two is thinking of a space as a topos (sheaves) or as a simplicial set (singular set). If a space has few non-constant functions into it, then singular cohomology is not 'strong enough'. $\endgroup$
    – David Roberts
    Commented May 30, 2011 at 0:30

4 Answers 4

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On seeing the question, I knew that I had seen something like this. Rinehart had a TAMS paper (Trans. Amer. Math. Soc. 174 (1972), 243-256) 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. :-)

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  • $\begingroup$ Tim, did you mean Appelgate, H., Tierney, M. (1969). Categories with models. In: Eckmann, B. (eds) Seminar on Triples and Categorical Homology Theory. Lecture Notes in Mathematics, vol 80. Springer, Berlin, Heidelberg. doi.org/10.1007/BFb0083086, when you mentioned Appelgate and Tierney? $\endgroup$
    – David Roberts
    Commented Mar 26 at 1:49
  • $\begingroup$ David, Yes, That's the one. $\endgroup$
    – Tim Porter
    Commented Mar 27 at 6:24
  • $\begingroup$ thanks for confirming $\endgroup$
    – David Roberts
    Commented Mar 27 at 6:36
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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.

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    $\begingroup$ There's easier sources than Bredon's book -- if I recall correctly (my copies are at my office and I'm home right now), there are nice proofs that singular and Cech cohomology agree in Griffiths and Harris and in Warner's book "Foundations of differentiable manifolds and Lie groups". Alternatively, it is pretty easy to verify that Cech cohomology satisfies the axioms for ordinary cohomology. $\endgroup$ Commented May 30, 2011 at 2:09
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    $\begingroup$ This is done also in Godement's book, if I recall correctly. $\endgroup$ Commented May 30, 2011 at 2:40
  • $\begingroup$ The beautiful book by Bott and Tu also covers the isomorphism between Cech and de Rham or singular cohomology. It is a very easy spectral sequence argument. Of course, one then still has to show that Cech cohomology agrees with derived functor cohomology. $\endgroup$ Commented May 30, 2011 at 20:05
  • $\begingroup$ Sorry - I didn't mean to imply the result is due to Bredon, just that I know it can be found there. $\endgroup$ Commented May 31, 2011 at 5:44
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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 :)

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    $\begingroup$ I guess if your functor were something like [X, K(G,n)] and your model structure were the standard model structure on Top, then you get cohomology as a Quillen derived functor, is that right..? $\endgroup$ Commented May 30, 2011 at 2:47
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    $\begingroup$ "Or perhaps not! If so, ask away." Yes, please! I'm interested in! $\endgroup$
    – user12832
    Commented May 30, 2011 at 6:16
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    $\begingroup$ @Daniel - not quite. Knowing that $H^n(X,G) = [X,K(G,n)]$ is a theorem (Brown representability). Or something something spectra. @unknown (google) - will try to put something in. $\endgroup$
    – David Roberts
    Commented May 30, 2011 at 9:38
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    $\begingroup$ +1 for "something something spectra" :) $\endgroup$ Commented May 30, 2011 at 16:54
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    $\begingroup$ right but if I understand this doesn't hold for a "bad" space e.g. unless X is a CW complex so I was hoping that by passing to a cofibrant replacement would fix this problem, but that was just a random idea... $\endgroup$ Commented May 30, 2011 at 22:15
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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.

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    $\begingroup$ @Saul I am afraid that can only be' is not a very exact statement! More strictly speaking abelianisation should refer to going from groups to abelian groups. The process you describe is more a linearisation' and even that is not a really good term. This becomes important in Quillen's work when he does a sort of relative abelianisation over an object in his paper on cohomology of algebras. This is the analogue of passing to the universal cover and taking homology, and takes account of the action of the fundamental group, hence `can only be' is not really correct. $\endgroup$
    – Tim Porter
    Commented May 31, 2011 at 6:27

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