Corollary 5.77.
Let \(K\) be a monogenic acyclic class. Then \(K^2 = K\).
Proof
The inclusion \(K^2 \subseteq K\) is immediate from Lemma 5.73. For the converse, we may assume that \(K = \Sigma^m\) for a class of monomorphisms \(\Sigma\), and it will suffice to show that \(u \in K^2\) for every morphism \(u\colon A \hookrightarrow B\) in \(\Sigma\). By construction, the morphism \(u \ssquare u\colon (B \times A) \sqcup_{A \times A} (A \times B) \to B \times B\) is in \(K^2\), hence so is its base change along \(\Delta\colon B \to B \times B\). Universality of pushouts identifies this base change with \(A \sqcup_{A\times_B A} A \to B\). Since \(u\) is a monomorphism, we have \(A \iso A \times_B A\), so this map identifies with \(u\colon A \to B\).