There exists a construction of the cohomology ring using only the Eilenberg–Steenrod axioms? I'm not able to find a reference where the theory is developed only with the axioms (I mean all of then, not generalized cohomology).

With construction, I mean the construction of the ring operation. With only the axioms I'm able to join all of the cohomology groups of a space in a direct sum but I don't know how to construct the multiplication. I know it comes from the diagonal map, but for this it would be necessary to prove the Kunneth theorem, and I don't know how to prove it from axioms. – Lucas Henrique Oct 17 '17 at 02:13
1 Answers
A ring structure isn't part of a generic generalized cohomology theory and its not derivable from the axioms.
One way in which it enters the picture would be for topological theories represented by ring spectra. Another, more algebraic way, would be for theories $E$ that have a Kunneth morphism $\mu:E^*(X)\otimes E^*(X)\rightarrow E^*(X\times X)$, but this is certainly not going to be true for all theories.
These conditions are in fact more or less the same as for spectra the Kunneth theorem is a special case of the universal coefficient theorem, which requires some sort of action maps between the involved spectra (Adams Stable Homotopy and Generalised Homology III.13).
In the case of singular cohomology one can lift the calculations to cochains and use the algebraic Kunneth theorems to derive the topological versions. You can see T. tom Dieck's Algebraic Topology 9.7 for a discussion of the EilenbergZilber theorem in this context.
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1But the question is asking about the EilenbergSteenrod axioms for _ordinary_ cohomology (with coefficients in a ring, presumably). One would hope it would be possible to construct the cup product just from the axioms, without using any particular construction. – Eric Wofsey Oct 17 '17 at 07:06

I'm sorry if I perhaps misunderstood the question then. I'll take down the answer if you think it isn't so relevant. Isn't the difference between ordinary and generalised cohomology contained in the dimension axiom? I'm not sure this is strong enough to give you a product. I'm following the reference *Foundations of Algebraic Topology* by Eilenberg and Steenrod. – Tyrone Oct 17 '17 at 07:29

1The dimension axiom plus assuming $H^0(*)$ is a commutative ring $R$ is enough to give a product, since it implies your cohomology theory is isomorphic to singular cohomology with coefficients in $R$. So the question is, can you construct this product directly from the axioms, without using a specific model of them like singular cohomology? – Eric Wofsey Oct 17 '17 at 07:37

For that to hold I think we need to either restrict the domain of definition to CW complexes or assume that the theory satisfies in addition the strong wedge axiom. – Tyrone Oct 17 '17 at 08:52

We could restrict to CW complexes and use a field as coefficients if it becomes easier. To put another axiom I think it would be unnecessary, because of the uniqueness of the theory. But I accept it gratefully if it let the construction possible. – Lucas Henrique Oct 17 '17 at 22:44