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...

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...

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...

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...