Does homology have a coproduct? - MathOverflow

Does homology have a coproduct?

2009-10-13T07:45:47Z

Standard algebraic topology defines the cup product which defines a ring structure on the cohomology of a topological space. This ring structure arises because cohomology is a contravariant functor and the pullback of the diagonal map induces the product (using the Kunneth formula for full generality, I think.)

I've always been mystified about why a dual structure, perhaps an analogous (but less conventional) "co-product", is never presented for homology. Does such a thing exist? If not, why not, and if so, is it such that the cohomology ring structure can be derived from it?

I am aware of the intersection products defined using Poincare duality, but I'm seeking a true dual to the general cup product, defined via homological algebra and valid for the all spaces with a cohomology ring.

Answer by Ben Webster for Does homology have a coproduct?

2009-10-13T12:56:22Z

Yes, homology is a coalgebra, at least with coefficients over a field, with the coproduct given by the pushforward X -> X x X. This is dual to the ring structure on cohomology, so they determine each other.

Answer by Andrew Stacey for Does homology have a coproduct?

2009-10-13T13:08:03Z

The Eilenberg-Zilber theorem says that for singular homology there is a natural chain homotopy equivalence:

S_*(X) ⊗ S_*(Y) ≅ S_*(X × Y)

The map in the reverse direction is the Alexander-Whitney map. Therefore we obtain a map

S_*(X) → S_*(X × X) → S_*(X) ⊗ S_*(X)

which makes S_*(X) into a coalgebra.

My source (Selick's Introduction to Homotopy Theory) then states that this gives H_*(X) the structure of a coalgebra. However, I think that the Kunneth formula goes the wrong way. The Kunneth formula says that there is a short exact sequence of abelian groups:

0 → H_*(C) ⊗ H_*(D) → H_*(C ⊗ D) → Tor(H_*(C), H_*(D)) → 0

(the astute will complain about a lack of coefficients. Add them in if that bothers you)

This is split, but not naturally, and when it is split it may not be split as modules over the coefficient ring. To make H_*(X) into a coalgebra we need that splitting map. That requires H_*(X) to be flat (in which case, I believe, it's an isomorphism).

That's quite a strong condition. In particular, it implies that cohomology is dual to homology.

Of course, if one works over a field then everything's fine, but then integral homology is so much more interesting than homology over a field.

In the situation for cohomology, only some of the directions are reversed, which means that the natural map is still from the tensor product of the cohomology groups to the cohomology of the product. Since the diagonal map now gets flipped, this is enough to define the ring structure on H^*(X).

There are deeper reasons, though. Cohomology is a representable functor, and its representing object is a ring object (okay, graded ring object) in the homotopy category. That's the real reason why H^*(X) is a ring (the Kunneth formula has nothing to do with defining this ring structure, by the way). It also means that cohomology operations (aka natural transformations) are, by the Yoneda lemma, much more accessible than the corresponding homology operations (I don't know of any detailed study of such).

Rings and algebras, being varieties of algebras (in the sense of universal or general algebra) are generally much easier to study than coalgebras. Whether this is more because we have a greater history and more experience, or whether they are inherently simpler is something I shall leave for another to answer. Certainly, I feel that I have a better idea of what a ring looks like than a coalgebra. One thing that makes life easier is that often spectral sequences are spectral sequences of rings, which makes them simpler to deal with - the more structure, the less room there is for things to get out of hand.

Added Later: One interesting thing about the coalgebra structure - when it exists - is that it is genuinely a coalgebra. There's no funny completions of the tensor product required. The comultiplication of a homology element is always a finite sum.

Two particularly good papers that are worth reading are the ones by Boardman, and Boardman, Johnson, and Wilson in the Handbook of Algebraic Topology. Although the focus is on operations of cohomology theories, the build-up is quite detailed and there's a lot about general properties of homology and cohomology theories there.

Added Even Later: One place where the coalgebra structure has been extremely successfully exploited is in the theory of cohomology cooperations. For a reasonably cohomology theory, the cooperations (which are homology groups of the representing spaces) are Hopf rings, which are algebra objects in the category of coalgebras.

Answer by Tyler Lawson for Does homology have a coproduct?

2009-10-15T11:58:14Z

Homology is not naturally a coalgebra unless you take field coefficients or unless your object has torsion free homology groups over the integers. The basic issue, as mentioned above, is that even though you have a split exact universal coefficient sequence for the homology of a product, the splitting isn't natural. You don't actually need homology to be dual to cohomology because that would involve some additional finiteness properties.

However, if your space has torsion-free homology with integer coefficients, then H_*(X;R) = H_*(X) ⊗ R for all R, and so you get a coalgebra structure on the homology of X with coefficients in R simply as the base change of the one over the integers. If R is an algebra over a field then you get a coalgebra structure with no assumptions on X by base-change from said field.

I should probably point out that the Kunneth formula is more complicated than stated in a previous answer. There's an exact sequence

0 → H_*(C;Z) ⊗ H_*(D;M) → H_*(C ⊗ D;M) → Tor(H_*(C;Z), H_*(D;M)) → 0

but notice that one side involves integer coefficients and the other coefficients in a general module. If you want the universal coefficient theorem with the same coefficients on both sides it takes the form of a spectral sequence with E_2-term

Tor^R_{p,q} (H_*(X;R), H_*(Y;R))

converging to H_*(X x Y;R). (The bigrading on Tor comes because we're taking Tor of graded modules.)

In general if E is a generalized homology-cohomology theory then flatness of E_* X over the ground ring E_* guarantees a coalgebra structure on the E-homology of X. This also may or may not have anything to do with duality, because flatness and projectiveness are not the same.

As mentioned, you still do have a coalgebra structure on the chains of X (or the "E-homology object of X" in the stable homotopy category), and this is really just some kind of failure of the homology groups to mimic what's going on behind the scenes.