Relationship between universal coefficient theorem and $[K(\mathbb{Z},n), K(G,n)]$?

Question

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 $H^n(X; G)$ we get when we're changing our coefficients from $\mathbb{Z}$ to $G$ (i.e., those that come simply from tensoring with the new group) corresponded to elements of $[K(\mathbb{Z}, n), K(G, n)]$. Then, the other classes that arise from $\operatorname{Ext} /\operatorname{Tor}$ would correspond to elements of $[X, K(G,n)]$ which don't factor through $K(\mathbb{Z},n)$. 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 $H^n(X; G)$ 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...

Answer 1

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 $H^n(K(\mathbb{Z}, n); G)$ coming from the universal class in $H^n(K(\mathbb{Z}, n); \mathbb{Z})$, and this will give your desired map $K(\mathbb{Z}, n) \to K(G, n)$. 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. $\operatorname{Ext}^{m-n}(G, H)$. 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 $\operatorname{Ext}$ 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 $\operatorname{Ext}$).

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 $n=m$.

Answer 2

This may not add much, but: If $0 \to \mathbb{Z}^r \to \mathbb{Z}^s \to G \to 0$ is a free resolution of your group, then you can construct a fibration sequence $K(\mathbb{Z}^s, n) \to K(G, n) \to K(\mathbb{Z}^r, n+1)$. 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

$$0 \to H^n(X; \mathbb{Z}) \otimes G \to H^n(X; G) \to \operatorname{Tor}(H^{n+1}(X; \mathbb{Z}), G) \to 0$$

(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 $[X, K(\mathbb{Z}^s, n)]$ up the fibration sequence, and the ones on the $\operatorname{Tor}$ side are what are left.

Answer 3

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.)