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

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

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

$$G \to G \times G,\;g \mapsto (g,1) \quad\text{ or }\quad G \to G \times G,\; g \mapsto (1,g)\;\; ? $$

onlya diagonal map on resolutions of $\mathbb Z$, and those can be taken non-coassociative and worse :-) ) – Mariano Suárez-Alvarez♦ Dec 16 '12 at 2:52uniquecomultiplication making that object into acounitalcoassociative coalgebra. So the answer to @TJ's question is that the other maps do not give a unital cup product, as can be seen explicitly from @John's answer, in which "cup product with $\beta$" is sometimes the zero map. – Theo Johnson-Freyd Dec 16 '12 at 5:10