Quotient triple factorization

(1) It is clear that \(\varphi^{\mono}\) is a monogenic quotient, that \(\varphi^{\quottext}\) is a quotient map, and that \(\varphi^{\cons}\) is conservative. Let \(q\colon T\to T/K^{\mono}\) denote the first quotient. The kernel of the induced quotient \(\varphi^{\epi}\colon T/K^{\mono}\to T/K\) is the congruence generated by the image \(q(K)\). By [Anel et al. 2024, Proposition 4.2.5], this image congruence is contained in the \(\infty\)-connected maps. It is therefore epigenic by Lemma 5.61. Consequently, \(\varphi^{\wcons}\) is weakly conservative.(2) Given a decomposition \(T \xrightarrow{\psi} T' \to S\) into a quotient map and a conservative map, we see that \(\ker(\psi) = \ker(\varphi) = K\). Since \(\psi\) is a quotient map, it follows that \(T' \iso T/K\).(3) Given a decomposition \(T \xrightarrow{\psi} T' \to S\) into a monogenic quotient map and a weakly conservative map, the fact that the second map only inverts effective epimorphisms implies Since \(\psi\) is a monogenic quotient, we have \(\ker(\psi)=\ker(\psi)^{\mono}=K^{\mono}\), and thus \(T'\iso T/K^{\mono}\).(4) This follows by first factoring \(\varphi\) as in (2) and then applying (3) to the quotient map \(T\to T/K\). These identifications also determine the comparison maps between any two such factorizations, proving uniqueness.