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](https://mathoverflow.net/questions/461/understanding-steenrod-squares), 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...



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


