# Chain map induces map of chain complexes and induced product, intuition behind isomorphism preserving products?

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?

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" $\mathbb{Z} \cong \mathbb{Z} \otimes \mathbb{Z}$.