Lemma 5.73.
We have an inclusion \(KL \subseteq K \cap L\).
Proof
By symmetry, it suffices to show \(KL \subseteq K\). If \(u \in K\) and \(v \in L\), then also the maps \(u \times C\) and \(u \times D\) are in \(K\), hence so is the cobase change \(A \times D \to (B \times C) \sqcup_{A \times C} (A \times D)\). By right cancellation, it follows that the cogap map \(u \ssquare v\) is in \(K\) as well.