Why do we use the diagonal for diagonal approximations ?

Question

First recall how the cup product is defined for the cohomology of a group $G$:

Fix a projective resolution $P \to \mathbb{Z}$ over $\mathbb{Z}G$. Then $P \otimes P \to \mathbb{Z} \otimes \mathbb{Z} = \mathbb{Z}$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G \otimes \mathbb{Z}G=\mathbb{Z}[G \times G]$. Since the diagonal $$D: G \to G \times G,\;g \mapsto (g,g)$$ is a group homomorphism, $P\otimes P$ can be considered as (acyclic) complex of $\mathbb{Z}G$-modules via $D$. By standard homological algebra there is a $\mathbb{Z}G$-linear map $\Delta: P \to P \otimes P$ (called a diagonal approximation) that extends $id: \mathbb{Z} \to \mathbb{Z}$. Finally, if $M,N$ are $\mathbb{Z}G$-modules, the cup product is defined on cochain level by the morphism

$$\begin{array}{lll} Hom_{\mathbb{Z}G}(P,M) \otimes Hom_{\mathbb{Z}G}(P,N) & \xrightarrow{} & Hom_{\mathbb{Z}(G\times G)}(P\otimes P,M\otimes N) \newline & \xrightarrow{\Delta^\ast} & Hom_{\mathbb{Z}G}(P,M\otimes N) \end{array}$$

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Obviously, the same construction can be made with any group homomorphism $G \to G \times G$ in place of $D$.

Question 1: What is the motivation to choose the diagonal $D$ for the definition of the cup product ?

Or, to put it the other way round:

Question 2: What "cup product" do be get if we choose one of the group homomorphisms
$$G \to G \times G,\;g \mapsto (g,1) \quad\text{ or }\quad G \to G \times G,\; g \mapsto (1,g)\;\; ? $$

Answer 1

The diagonal map $\Delta$ is "coassociative": the two maps $(\Delta \otimes 1) \circ \Delta$ and $(1 \otimes \Delta) \circ \Delta$ from $\mathbb{Z}G$ to $(\mathbb{Z}G)^{\otimes 3}$ are equal. Therefore $\Delta$ induces an associative product on cohomology. Similarly, the map $\varepsilon: \mathbb{Z}G \rightarrow \mathbb{Z}$ defined by $\varepsilon(g)=1$ for all $g \in G$ induces the unit map $\mathbb{Z} \to H^*(G)$ in cohomology for the product induced by $\Delta$, because of how $\Delta$ and $\varepsilon$ interact. Other choices for "diagonals" won't do this, in general.

(The fancy thing to say is that $\mathbb{Z}G$ is a Hopf algebra, with structure maps given by the usual product and unit, along with $\Delta$ and $\varepsilon$, and also a map $\chi: \mathbb{Z}G \to \mathbb{Z}G$ defined by $\chi(g) = g^{-1}$ for $g \in G$.)

For question 2, Dylan is saying that the map $g \mapsto (g,1)$ will induce the "product" map $\alpha \otimes \beta \mapsto \alpha \beta$ if $\beta \in H^0(G)$ (i.e. $\beta$ is a scalar), $\alpha \otimes \beta \mapsto 0$ otherwise.