If $\Gamma=\Gamma_1\sqcup\Gamma_2$ is a nontrivial <disjoint union of graphs>, no defining relation mixes the two vertex sets, and hence
$$
A_\Gamma\cong A_{\Gamma_1}*A_{\Gamma_2}.
$$
On the topological side, every clique lies in one component, so
$$
S_\Gamma=S_{\Gamma_1}\vee S_{\Gamma_2},
$$
a one-point union of <Salvetti complexes>.

If $\Gamma=\Gamma_1*\Gamma_2$ is a nontrivial <join of graphs>, every generator from the first part commutes with every generator from the second. Therefore
$$
A_\Gamma\cong A_{\Gamma_1}\times A_{\Gamma_2}.
$$
Every clique of the join is the union of a clique in each factor, which gives the cubical identity
$$
S_\Gamma\cong S_{\Gamma_1}\times S_{\Gamma_2}.
$$

Solution

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

{wiki=Join_(graph_theory)}

The join $\Gamma_1*\Gamma_2$ is obtained from the <disjoint union of graphs> $\Gamma_1\sqcup\Gamma_2$ by adding every edge between a vertex of $\Gamma_1$ and a vertex of $\Gamma_2$.

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