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Michael Albanese
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This may not add much, but: If 0 -> Zr -> Zs -> G -> 0$0 \to \mathbb{Z}^r \to \mathbb{Z}^s \to G \to 0$ is a free resolution of your group, then you can construct a fibration sequence K(Zs,n) -> K(G,n) -> K(Zr,n+1)$K(\mathbb{Z}^s, n) \to K(G, n) \to K(\mathbb{Z}^r, n+1)$. 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 -> Hn(X;Z) ⊗ G -> Hn(X;G) -> Tor(Hn+1(X;Z),G) -> 0

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

(and I'm worried that maybe G$G$ has to be finitely generated). TheThe cohomology classes on the tensor side come from those maps that lift from [X,K(G,n)]$[X, K(G, n)]$ to [X,K(Zs,n)]$[X, K(\mathbb{Z}^s, n)]$ up the fibration sequence, and the ones on the Tor$\operatorname{Tor}$ side are what are left.

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.

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.

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Tyler Lawson
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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.