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I've looked around on MSE and online only to find scattered results, which confuse me. I want to understand how homology behaves with (co)limits. I want to know in particular about singular homology, and in general about homology theories.

  1. Does singular homology commute with arbitrary colimits? Where can I find proof of this? Under what conditions does it behave well with limits?
  2. In A Concise Course in Algebraic Topology, the author proves using the additivity, weak equivalence, and MVS axioms that homology preserves direct limits. Unless I'm mistaken, preserving direct limits is equivalent to preserving filtered colimits. Is this correct? Is there a simpler proof for May's result?

Added: In particular, I am confused by the answer to this question according to which homology preserves all kinds of colimits.

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    Read carefully! The answer says homology (of chain complexes) preserves filtered colimits. But singular homology (of topological spaces) probably only preserves coproducts. – Zhen Lin Feb 18 '15 at 20:58
  • @ZhenLin regarding the answer - seeing how it emphasizes $\pi _n$ preserve filtered colimits, but does not mention the word filtered for homology, I thought the author meant arbitrary colimits. Regarding singular homology - isn't the singular functor left adjoint, thereby preserving arbitrary colimits? Wouldn't this make the composition - the singular homology functor - also preserve filtered colimits? –  Feb 18 '15 at 21:07
  • @Exterior: the singular simplicial set functor is a right adjoint. A priori it preserves no colimits whatsoever. – Qiaochu Yuan Feb 18 '15 at 21:17
  • @QiaochuYuan ah, right. Thanks. –  Feb 18 '15 at 21:43
  • @Exterior (I believe that Čech homology, or Steenrod-Sitnikov homology, have better properties wrt. (co)limits, but am not an expert) – Peter Franek Feb 18 '15 at 21:56
  • @PeterFranek I think those are far beyond me for now. Hope to get to them eventually :D –  Feb 18 '15 at 23:04
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    Off-topic, but... When you search MSE, what would you expect if not scattered results? – darij grinberg Feb 18 '15 at 23:56

2 Answers2

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Singular homology already fails to commute with pushouts. A pushout of spaces doesn't give a pushout of homology groups, but instead gives (maybe under niceness conditions) a long exact sequence. For an explicit example consider the pushout of

$$D^2 \leftarrow S^1 \rightarrow D^2$$

which is $S^2$. This doesn't induce a pushout on $H_2$.

Singular homology also fails to commute with products. (Note that the tensor product is not the product in the category of abelian groups, or of graded abelian groups, so even if we're working over a field the Kunneth formula is not a response to this claim.)

The first fact is in some sense a reflection of a failure to be suitably higher categorical. There is a very abstract description of what it means to compute the homology (not the homology groups, but "the homology") of a space, namely tensoring it with some spectrum, and this construction preserves all homotopy colimits (in fact it is a left adjoint in a higher categorical sense). It's very natural to think about homotopy colimits rather than colimits because taking singular homology is homotopy-invariant, but taking colimits is not, while taking homotopy colimits is.

Then you have to figure out how to compute homotopy colimits of spaces, and also figure out what a homotopy colimit of spectra buys you once you pass to homotopy groups (e.g. long exact sequences, or more generally spectral sequences).

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  • Thanks for the interesting answer! I'm still confused by what I see in May's book: on page 115 I think he shows homology $E_{\ast}:\mathsf{Top}^2 \rightarrow \mathsf{Ab}$ preserves some kind of "linearly ordered" colimit". What am I misunderstanding? –  Feb 19 '15 at 10:21
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    I think I understand what I was missing: Since May requires the diagram to be linearly ordered, there's no contradiction with not preserving pushouts. However, Zhen Lin mentioned before that singular homology does commute with coproducts. This does not follow from preserving "linearly ordered colimits" right? So preserving coproducts just comes from a separate argument? Or is this the reason for the additivity axiom? –  Feb 19 '15 at 15:02
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    @Exterior: right, there's no contradiction. I would be tentatively willing to believe that singular homology preserves filtered colimits but I don't have a proof off the top of my head. Preserving coproducts is separate, and there's no contradiction with not preserving pushouts because those aren't filtered. – Qiaochu Yuan Feb 19 '15 at 15:06
  • I'm looking at prop 23.1 here (chain homology commutes with filtered colimits), but I'm confused: the author starts by supposing we have something in $H(\varinjlim F)$, and then says such an element is represented by an $n$-cycle in $(\varinjlim F)_n$ for some $n$. But $H(\varinjlim F)$ is a graded abelian group, so the previous sentence makes no sense to me. Could you please clarify? –  Feb 20 '15 at 15:17
  • @Exterior: I don't understand the question. By definition, a class in the homology of a chain complex $H(C)$ is represented by an $n$-cycle in $C_n$ for some $n$. – Qiaochu Yuan Feb 21 '15 at 04:46
  • Oh, I never saw this terminology when talking about the entire homology at once. So by "a class in the homology $H(C_\bullet)$" one actually means an element in one of the homology objects $H_n(C_\bullet)$? Finally, the proposition only shows a bijection, so to be formal I'd have to mention the forgetful functor $U:\mathsf{Ab}\rightarrow \mathsf{Set}$ reflects filtered colimits, right? –  Feb 21 '15 at 16:08
  • @Exterior: right. – Qiaochu Yuan Feb 21 '15 at 19:17
  • @Exterior In general, chain homology functor commutes with exact functors, a consequence of the FHHF theorem (cf. Ravi Vakil's notes, FOAGjan2915, 1.6.H). If, instead of exact functors, consider arbitrary right exact functor, then maybe a spectral sequence intervenes. – Yai0Phah Mar 07 '15 at 08:32
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This is not exactly an answer to your question, except that it is relevant to the question of homotopical invariants preserving at least some colimits. For example, the fundamental group of based spaces does not preserve all colimits: the usual Seifert-van Kampen Theorem determines the fundamental group of a union $X=U \cup V$ of based spaces if $U,V$ are open and $U \cap V$ is path connected. See this mathoverflow discussion on this issue. .

A generalisation of this theorem, dating to 1981, and also involving connectivity conditions, is given in the 2011 book Nonabelian Algebraic Topology, and concerns in part a homotopically defined functor $\Pi$ from filtered spaces to "crossed complexes", which are a kind of partially nonabelian chain complex with operators. This work does not involve an input from singular homology, although the relation to that is of interest.

Background to this work is given in this recent paper Modelling and Computing Homotopy Types: I, which gives a methodology for the use of colimit arguments in homotopy theory.

We also know that that homology groups as functors from chain complexes to say graded abelian groups do not preserve colimits.

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