Lemma 5.74.
Given two classes of maps \(\Sigma,\Sigma'\) in \(T\), we have \(\Sigma^m\,{\Sigma'}^m \;=\; (\Sigma \ssquare \Sigma')^m\). In particular, if \(K\) and \(L\) are acyclic classes of small generation, then so is \(KL\).
Proof
The inclusion \((\Sigma \ssquare \Sigma')^m \subseteq \Sigma^m \, {\Sigma'}^m\) is clear. For the converse, it suffices to show that \((\Sigma \ssquare \Sigma')^m\) contains \(\Sigma^m \ssquare {\Sigma'}^m\). Let us first show that it contains \(\Sigma \ssquare {\Sigma'}^m\). For a fixed morphism \(u \in \Sigma\), consider the collection of morphisms \(v \in \Ar(T)\) such that \(u \ssquare v \in (\Sigma \ssquare \Sigma')^m\). This collection clearly contains \(\Sigma'\), hence to show it contains all of \({\Sigma'}^m\) it remains to show that it is a saturated class closed under base change.Closure under base change follows from universality of colimits: if \(v'\) is a base change of \(v\), then \(u\ssquare v'\) is the corresponding base change of \(u\ssquare v\). The class clearly contains all isomorphisms. For closure under composition, let \(C\xrightarrow{v}D\xrightarrow{w}E\) be composable. The map \(u\ssquare(wv)\) factors as
\[(B\times C)\sqcup_{A\times C}(A\times E)
\longrightarrow
(B\times D)\sqcup_{A\times D}(A\times E)
\longrightarrow B\times E.\]
The first map is a cobase change of \(u\ssquare v\), and the second is \(u\ssquare w\). Thus it belongs to \((\Sigma\ssquare\Sigma')^m\) whenever both \(u\ssquare v\) and \(u\ssquare w\) do. Finally, closure under colimits follows because \(u\ssquare-\colon\Ar(T)\to\Ar(T)\) preserves colimits.We may now repeat the argument to show that for fixed \(v \in {\Sigma'}^m\) the collection of \(u \in \Ar(T)\) such that \(u \ssquare v \in (\Sigma \ssquare \Sigma')^m\) contains \(\Sigma\) and is a saturated class closed under base change, hence contains all of \(\Sigma^m\). This finishes the proof.