Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 134 / 1 / b / Solution

If $\Gamma ={\Gamma }_1\bigsqcup {\Gamma }_2$ is a nontrivial disjoint union of graphs, no defining relation mixes the two vertex sets, and henceOn the topological side, every clique lies in one component, soa one-point union of Salvetti complexes.

If $\Gamma ={\Gamma }_1\ast {\Gamma }_2$ is a nontrivial join of graphs, every generator from the first part commutes with every generator from the second. ThereforeEvery clique of the join is the union of a clique in each factor, which gives the cubical identity