# relationship between universal coefficient theorem and [K(Z,n),K(G,n)]?

In short, I'm wondering whether the universal coefficient theorem can be understood/reinterpreted by using maps of Eilenberg-MacLane spaces. This is a wishy-washy idea and I don't have evidence to back it up, but it would be very nice if the "freebie" cohomology classes in Hn(X;G) we get when we're changing our coefficients from ℤ to G (i.e., those that come simply from tensoring with the new group) corresponded to elements of [K(ℤ,n),K(G,n)]. Then, the other classes that arise from Ext/Tor would correspond to elements of [X,K(G,n)] which don't factor through K(ℤ,n). Is anything like this even remotely true?

This question is in part motivated by the responses to an earlier question of mine, which mentioned that viewing Hn(X;G) as [X,K(G,n)] helps us understand cohomology operations (in that case, Steenrod squaring). It seems as if the representability of cohomology is probably only useful for studying honest cohomology operations, but I don't think I understand exactly what that means well enough to deduce whether changing coefficients qualifies...

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You can indeed! By Yoneda, any natural way of getting cohomology classes can be realized on the Eilenberg-MacLane spaces. You can look at those "freebie" classes in H^n(K(Z,n);G) coming from the universal class in H^n(K(Z,n);Z), and this will give your desired map K(Z,n) \to K(G,n). However, there's a nice bigger story behind this specific case besides just Yoneda.

First, the Dold-Kan theorem says that (nonnegatively graded) chain complexes of abelian groups are equivalent to simplicial abelian groups. More precisely, given a simplicial abelian group, the alternating sum of the face maps turns it into a chain complex, and modding out by the image of the degeneracies gives the "normalized" chain complex. This turns out the be an equivalence of categories (the inverse is taking a chain complex and formally adding degenerate things to it). What's more, this equivalence preserves the usual notion of homotopy in the two categories: the derived category of abelian groups is the same as simplicial abelian groups with weak equivalences formally inverted.

Now K(G,n) can be realized as the simplicial abelian group which corresponds to the chain complex with G in degree n and 0 everywhere else. Maps of simplicial abelian groups (mod homotopy) from K(G,n) to K(H,m) are then the same as maps of chain complexes in the derived category, i.e. Ext^{m-n}(G,H). This is 0 except for m=n and m=n+1, and in those cases you get exactly the correspondence that the universal coefficient theorem gives you. The cohomology operations with m=n+1 (coming from the Ext part of the universal coefficient theorem) are called Bocksteins and can also be obtained as the connecting homomorphisms of a long exact sequence on cohomology coming from a short exact sequence of coefficient groups (namely, the group extension corresponding to your element of Ext).

Note that there are lots of other operations on cohomology (eg, Steenrod operations) besides these. What makes these special is that they can be implemented by maps from K(G,n) to K(H,m) which are group homomorphisms with respect to the abelian group structures on the spaces. Note, however, that any cohomology operation which commutes with addition (which includes all Steenrod operations) is a group homomorphism up to homotopy, since the group structure on cohomology is just the group structure on K(G,n) taken modulo homotopy. Nevertheless, it is impossible to straighten these out to be homomorphisms on the nose, except in the case of Bocksteins and in the case n=m.

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This may not add much, but: If 0 -> Zr -> Zs -> G -> 0 is a free resolution of your group, then you can construct a fibration sequence K(Zs,n) -> K(G,n) -> K(Zr,n+1). You can then get something like the "factorizations" you're trying to interpret from the universal coefficient theorem that way, but it's from a somewhat less-standard universal coefficient theorem for cohomology

0 -> Hn(X;Z) ⊗ G -> Hn(X;G) -> Tor(Hn+1(X;Z),G) -> 0


(and I'm worried that maybe G has to be finitely generated). The cohomology classes on the tensor side come from those maps that lift from [X,K(G,n)] to [X,K(Zs,n)] up the fibration sequence, and the ones on the Tor side are what are left.

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Just a minor addition to the other answers, since you seem particularly interested in cohomology theories and operations on them. We often concentrate on operations of a particular cohomology theory, but the theory works equally well for operations between cohomology theories. What you describe is a particular instance of that. Where the cohomology theories are particularly well-behaved, all this operational structure is a sort-of souped-up version of rings and modules. You can think of the operations of a single cohomology theory as like a ring, and operations between cohomology theories as like a bimodule. (This is an analogy, they aren't rings and bimodules, they're Tall-Wraith monoids and bimodules for such.)

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