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(I apologize that this is a vague question).

I seems to me somehow that homotopy groups behave well with respect to (Serre)-fibrations. For example you get a long exact sequence of homotopy groups from it. On the other hand cofibrations and homotopy groups seems to be no good friends at all (e.g. $S^1\to D^2\to S^2$).

But then again, the situation in homology seems to be the other way round. They behave well with respect to cofibrations (you get a long exact sequence) and fibrations are harder to investigate (Serre spectral sequence etc.).

My question is: What is the intuition behind this difference? (in particular with respect to the fact that homology is just homotopy of another space.)

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up vote 21 down vote accepted

I agree that the long exact sequence in homotopy groups of a fibration follows from the fact that fibrations are defined using a mapping property in which the fibration is the target.

One way to understand why homology behaves well with respect to cofibrations is to spell out your remark that "homology is just homotopy of another space". This is true, but not obvious. There are a number of constructions of ordinary homology which take the following form.

One finds a functor $F$ from (pointed) spaces to (pointed) spaces which takes cofibrations to quasifibrations. (A quasifibration is something for which you have a long exact sequence of homotopy groups, for example a Serre fibration). And then $H_* (X) \cong \pi_* F(X).$ If $X\to Y \to Z$ is a cofibration (maybe a cofibration of CW complexes), then $\dots \to \pi_* F(X) \to \pi_* F(Y) \to \pi_* F(Z) \to \ldots$ is the long exact sequence in homology associated to the cofibration.

Here are several contexts in which one can describe such a functor $F$.

First, a formal approach. Let $\mathbf{S}$ denote the category of spectra: it is connected to the category $\mathbf{T}$ of spaces by adjoint functors $\Sigma^\infty: \mathbf{T} \to \mathbf{S}$ and $\Omega^\infty: \mathbf{S} \to \mathbf{T}.$ There a spectrum called the "Eilenberg-MacLane" spectrum, denoted $H\mathbb{Z}$: its job is to represent singular cohomology, and one can take $F(X) = \Omega^\infty ((\Sigma^\infty X) \wedge H\mathbb{Z})$.

Why does $F$ have the cofibration-to-quasifibration property? Well, the way that this is set up, $\Sigma^\infty$ preserves cofibrations of CW complexes, and $\Omega^\infty$ preserves fibrations of fibrant objects, and in the category of spectra every cofibration is equivalent to a fibration.

To be more explicit about $H\mathbb{Z}$, you can define $F(X) = \lim (\ldots \Omega^k(X \wedge K(\mathbb{Z},k)) \to \Omega^{k+1} (X\wedge K(\mathbb{Z},k+1))a \ldots),$ where the limit is a colimit and the maps defining the system arise from the maps $\Omega K(\mathbb{Z},k) \simeq \Omega K(\mathbb{Z},k+1)$.

Second, the Dold-Thom theorem says that one can take $F(X) = Sp^\infty (X).$ Here $Sp^n(X) = X^n/\Sigma_n$, and $Sp^\infty(X) = \lim \ldots Sp^n(X) \to Sp^{n+1}(X) \ldots$, again the limit is a colimit.

Third, if you're willing to allow $X$ to be a simplicial set, then one can take then one can take $F(X) = \mathbb{Z}X$, the simplicial set whose $n$ simplices are the free abelian group on the $n$-simplices of $X$. (This approach is due to Dan Kan; see the proceedings of the Hurewicz conference)

All of this is to focus attention on functors which take cofibrations to quasifibrations. In fact all $-1$-connected generalized homology theories (at least the ones associated to cohomology theories: are there homology theories which are not? I don't know) are of the form $E_* X = \pi_* G(X)$, where $G$ is a functor which takes cofibrations to quasifibrations. Indeed one takes $G(X) = \Omega^\infty (\Sigma^\infty X \wedge R)$, where $R$ is the spectrum representing the cohomology theory. This approach goes back to G. W. Whitehead.

One of the more compact discussions of such a functor, which I like, is in an article by G. Segal in Springer LNM 575; he gives a construction of connective real $K$-homology there. Really he's showing how to generalize the work of Dold and Thom: Segal's argument applies just as well to $Sp^\infty(X)$.

I apologize that throughout I have done a poor job of saying how to handle basepoints.

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Fibrations are defined as target-type concepts -- they have good formal homotopy properties when you map into them, for example, if you apply the functor $\pi_n(-) = [S^n,-]$.

Now I'll subject you to my point of view on homology.

Dually, cofibrations are a domain-type concept, so they behave well when you map out of them, say if you apply (represented) cohomology: $\widetilde H^n(-;G) = [-, K(G,n)]$.

Homology is a bit of a monster: it is a covariant functor that works well with domain-type input. It is a small miracle that such functors exist at all; this is why they are hard to construct. But the answer for homology has to be: homology works well with cofibrations because we built it to work well with cofibrations, and a more informative answer to your question would depend on the construction you use.

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@Jeff: I don't get this: "Homology is a bit of a monster: it is a covariant functor that works well with domain-type input. It is a small miracle that such functors exist at all; this is why they are hard to construct." – John Klein Feb 24 '11 at 21:35
@John: As far as I know, the exactness of homology (or its invariance with respect to suspension) depends on a substantial theorem (such as Blakers-Massey, or Freudenthal, or something else nontrivial). There is no way that I know of to construct a functor that is covariant and exact on cofibrations without invoking a real theorem. – Jeff Strom Feb 24 '11 at 22:47
What about the ordinary definition of singular homology? – Eric Wofsey Feb 25 '11 at 0:13
Eric, having spent an hour yesterday putting the proof of homotopy invariance on the blackboard in class, I'm going to agree with Jeff: it's a major pain! Yes, there's a nice geometric idea about prisms, and Hatcher's rather loose notation makes it look a little simpler than it really is, but the details are messy and not at all formal. For instance, how did Hatcher decide to break up the sums the way he did in his proof? One has to get one's hands pretty dirty before coming away with a clean argument. There's hard work in there! Same goes for barycentric subdivision of chains. – Dan Ramras Feb 26 '11 at 2:10
I know this comment thread is long dead, and it's subjective to claim something is "easy" or "hard". But maybe this has to do with Jeff's opinions: $\pi_\bullet$ and $H^\bullet$ are (co)representable functors. Yet homology involves smashing with something, and smash product is surely a complex idea. Of course, passing to a spectrum-level discussion is already absurd in a conversation about "easiness", but it certainly shows you don't get the properties of homology "for free", the way you do for (co)representable functors. – Hiro Lee Tanaka Sep 25 '13 at 4:10

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