Lemma 5.29.
Let \(L\) be a modality, and consider the zigzag construction associated to a span \(Q_0 \leftarrow P_0 \to R_0\) over \(B \leftarrow A \to C\). If \(P_{n-1} \to P_n\) lies in \(L\) for some \(n\geq 2\), then also \(P_n \to P_{n+1}\) lies in \(L\).
Proof
Suppose first that \(P_n\) is obtained from \(P_{n-1}\) by making the left leg cartesian, so that the next step makes the right leg cartesian. Since \(n\geq 2\), the preceding step made the right leg cartesian, so
\[P_{n-1} \iso R_{n-1}\times_C A.\]
Unwinding the definitions of these two consecutive steps gives \[R_n \iso P_n \sqcup_{P_{n-1}} R_{n-1}
\qquad\text{and}\qquad
P_{n+1} \iso R_n\times_C A.\]
Since pushouts are universal, it follows that \[P_{n+1}
\iso (P_n\times_C A)\sqcup_{P_{n-1}\times_C A}(R_{n-1}\times_C A)
\iso (P_n\times_C A)\sqcup_{P_{n-1}\times_C A}P_{n-1}.\]
Under this identification, the composite \(P_{n-1}\to P_n\to P_{n+1}\) is the canonical map from the second summand. It is therefore the pushout of \[P_{n-1}\times_C A \longrightarrow P_n\times_C A\]
along \(P_{n-1}\times_C A\to P_{n-1}\). The displayed map is a base change of \(P_{n-1}\to P_n\), hence lies in \(L\). It follows that the composite \(P_{n-1}\to P_{n+1}\) lies in \(L\). Since \(P_{n-1}\to P_n\) lies in \(L\) by assumption, right cancellation for the class \(L\) now implies that \(P_n\to P_{n+1}\) lies in \(L\).The case where \(P_n\) is obtained by making the right leg cartesian is symmetric, with \(B\) in place of \(C\).