I know basically the definition and very general-not-so-useful properties of derived categories, and to build a deeper understanding of them, I'd like to see if it can help in re-thinking some basic results in homological algebra.

More specifically, I'd like to know if the theory of derived categories can help in proving (perhaps some improvements of) the following : the universal coefficients formulas (homological with $\otimes$ and cohomological with $\hom$) and/or the (algebraic) Künneth formula.

It seems reasonable enough that there should be more conceptual proofs of these (or generalizations) through derived categories, but I haven't found one (which indicates either that it doesn't exist, or is a good witness of my lack of practice).

Perhaps there would be a way to get spectral sequences from an abstract "derived" argument, to get to a more concrete thing that would be one of these theorems ? If so I'd like to see that as well, as I still haven't quite seen how to get spectral sequences from derived stuff (and it seems to be a general theme)


The derived category of abelian groups is somewhat special that makes the Künneth and universal coefficient theorems take an unusually special form.

An abstract way to state this property is

Theorem: Every element of the derived category of abelian groups is the direct sum of one-term complexes

Proof: Every chain complex is quasi-isomorphic to a complex of free abelian groups. And if $C_\bullet$ is a complex of free abelian groups, the fact every subgroup of a free abelian group is free implies you can decompose $C_n = \ker(\partial_n) \oplus \mathrm{im}(\partial_n)$, and thus you can decompose $C_n$ into a direct sum of complexes of the form $\mathrm{im(\partial_n)} \to \ker(\partial_{n-1})$, each of which is isomorphic to the one-term complex $H_{n-1}(C_\bullet)$ concentrated in degree $n-1$. $\square$

In particular, the equivalence class of every chain complex $C_\bullet$ includes the complex

$$ \ldots \xrightarrow{0} H_1(C_\bullet) \xrightarrow{0} \underline{H_0(C_\bullet)} \xrightarrow{0} H_{-1}(C_\bullet) \xrightarrow{0} \ldots $$

which, of course, breaks apart into the direct sum of its individual terms.

From the form of tor and ext for one-term complexes, we can then write

$$ H_n (C_\bullet \otimes^\mathbb{L} D_\bullet) \cong H_{n-i-j} \left( \bigoplus_i \bigoplus_j H_i(C_\bullet) \otimes^\mathbb{L} H_j(D_\bullet) \right) \\ \cong \bigoplus_i \bigoplus_j \begin{cases} H_i(C_\bullet) \otimes H_j(D_\bullet) & n = i+j \\ \mathrm{tor}(H_i(C_\bullet), H_j(D_\bullet)) & n = i+j +1 \end{cases}$$

The universal coefficient theorem is the special case where $C_\bullet$ is the complex of coefficients concentrated in degree zero. Similarly,

$$ H_n (\mathbb{R}{\hom}(C_\bullet, D_\bullet)) \cong H_{n+i-j} \left(\prod_i \bigoplus_j \mathbb{R}{\hom}(C_i, D_j) \right) \\ \cong \prod_i \bigoplus_j \begin{cases} \hom(H_i(C_\bullet), H_j(D_\bullet)) & n = j-i \\ \mathrm{ext}(H_i(C_\bullet), H_j(D_\bullet)) & n = j-i-1 \end{cases} \\\cong \prod_i \hom(H_i(C_\bullet), H_{n+i}(D_\bullet)) \oplus \mathrm{ext}(H_i(C_\bullet), H_{n+i+1}(D_\bullet)) $$

With $D$ concentrated in degree zero, this becomes the familiar

$$ H_{-n} (\mathbb{R}{\hom}(C_\bullet, D)) \cong \hom(H_n(C_\bullet), D) \oplus \mathrm{ext}(H_{n-1}(C_\bullet), D) $$

  • $\begingroup$ Thank you for your answer ! I have a few questions about it : 1- you said abelian groups but I'm assuming this works pretty much for any PID and modules over it ? 2- would you know where I can find a proof of the theorem ? 3- If I'm not mistaken, there is a Künneth spectral sequence even over non PID's : can it be explained with similar ideas ? Thanks again ! $\endgroup$ – Max Sep 8 at 13:13
  • $\begingroup$ @Max: I've added a brief proof, which will indicate where I've used the the fact I'm working over abelian groups. I don't know off hand how general the result is. $\endgroup$ – Hurkyl Sep 8 at 20:54
  • $\begingroup$ I haven't thought too hard about how the general Künneth theorems would be formulated; I was sort of using this question as motivation to work through some things that I thought were true but hadn't gotten around to working through the details, namely this way in which Ab is special, and how to show these short exact sequences are split without having to dive into the fine detail. $\endgroup$ – Hurkyl Sep 8 at 20:57
  • $\begingroup$ Thank you for your additions. Your proof works fine for PIDs as I expected. One negative aspect of your answer (although it is very interesting !) is that you get an isomorphism with the direct sum instead of "there is a nonnaturally split natural exact sequence". I'll wait and see if someone has any addition concerning spectral sequences or other examples - if no one does I'll award you the bounty $\endgroup$ – Max Sep 8 at 21:09

I have not read it too much, but in the beggining (1.3) of this the author relates derived stuff, Künneth and spectral sequences. Thus it seems that at least this short paragraph might be something that you are searching for or at least some indication of it (not sure about the rest of the document).

  • $\begingroup$ Thank you ! The document does state the sort of derived analogue I was looking for (and gives a reference, Verdier's thesis, which fortunately is easily accessible - freely, on the site of the French mathematical society) I will wait a bit before accepting your answer, because I'd like to see if someone has another reference with proofs for instance $\endgroup$ – Max Jul 7 at 22:19
  • $\begingroup$ Glad that I could help. Sure, no worries. $\endgroup$ – ThorWittich Jul 7 at 22:31

I think that "homological functors takes shorts exact sequences to long exact sequences" is a sentence that is better written un triangulated categories. In fact, the definition of a triangle is a strange mix of the notions of short/long exact sequence. In my personal opinion, the notion of triangle is precisely the "theorem" or " principle" relating them. If you want a precise theorem, take "in a triangulated category, a representable functor is homological".


Since you asked for spectral sequences: the universal coefficient theorem is a special case of the change-of-rings spectral sequence.

Given a chain complex $C$ of $R$-modules of homological type, zero in negative degrees (the homological type is a convenience, but bounded below is necessary) and an $R$-algebra $S$, we have a homological first quadrant spectral sequence with $E^2$-term $$ E^2_{pq} = \mathrm{Tor}_p^R(S, H_q(C)) \implies H_{p+q}( S \otimes C).$$ When $R$ is a PID, then $\mathrm{Tor}_q^R = 0$ for $q \geq 2$, and the resulting spectral sequence collapses at $E^2$, yielding the familiar short exact sequences. This is essentially equivalent to the calculation in Hurkyl's answer.

In derived categorical language, this spectral sequence corresponds to computing the homology of the derived tensor product $S \otimes_R^{\mathbb L} C$ in terms of the derived tensor product of the homologies. This translates into spectral sequences: we can take a resolution of $C$ by a double complex of projective objects, tensor with $S$, and take spectral sequences associated to this double complex.

The Künneth formula over general $R$ is the same: there is a spectral sequence $$E^2_{pq} = \bigoplus_{q_1 + q_2 = q} \mathrm{Tor}^R_p (H_{q_1}(C_1), H_{q_2}(C_2)) \implies H_{p+q}(C_1 \otimes C_2).$$

  • $\begingroup$ Are you saying that this is the way people get spectral sequences out of derived things, is there no more general argument that allows to say "we have this situation in the derived world, and thus it yields a spectral sequence" ? (>+it's not super clear how you get the general Künneth spectral sequence from what you said as we have 2 complexes we have to resolve by double complexes) $\endgroup$ – Max Sep 11 at 8:05

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