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I know there is an isomorphism$$H^*(K(\pi, 1), A) \cong \text{Ext}_{\mathbb{Z}[\pi]}^*(\mathbb{Z}, A).$$When $A$ is a commutative ring, the $\text{Ext}$ groups have algebraically defined products, constructed as follows. The evident isomorphism $\mathbb{Z} \cong \mathbb{Z} \otimes \mathbb{Z}$ is covered by a map of free $\mathbb{Z}[\pi]$-resolutions $P \to P \otimes P$, where $\mathbb{Z}[\pi]$ acts diagonally on tensor products, $\alpha(x \otimes y) = \alpha x \otimes \alpha y$. This chain map is unique up to chain homotopy. It induces a map of chain complexes$$\text{Hom}_{\mathbb{Z}[\pi]}(P, A) \otimes \text{Hom}_{\mathbb{Z}[\pi]}(P, A) \to \text{Hom}_{\mathbb{Z}[\pi]}(P, A)$$and therefore an induced product on $\text{Ext}_{\mathbb{Z}[\pi]}^*(\mathbb{Z}, A)$. My question is, what is the intuition behind the isomorphism above preserving products?

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Intuitively, the isomorphism preserves products because just as the cup product comes from pulling back Künneth via the diagonal, the product seems to come from pulling back, i.e. dualizing via $\text{Hom}$, a derived version of the "Künneth" $\mathbb{Z} \cong \mathbb{Z} \otimes \mathbb{Z}$.

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