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.