Let \(C\) be a category with pullbacks such that groupoid colimits are universal in \(C\) (e.g. \(C\) is a topos).
A morphism \(f\colon U \to X\) in \(C\) is an effective epimorphism if and only if the functor \(f^*\colon C_{/X} \to C_{/U}\) is conservative.
Effective epimorphisms are closed under base change.
Proof
(1) Assume first that \(f\) is an effective epimorphism. By Lemma 2.26, \(X \simeq \colim_n \check{C}_n(f)\) is a groupoid colimit. By our assumption on \(C\), the functor \(C_{/X} \hookrightarrow \lim_n C_{/\check{C}_n(f)}\) is fully faithful, hence conservative. In particular, the functors \(C_{/X} \to C_{/\check{C}_n(f)}\) for \([n] \in \simp\catop\) are jointly conservative. Since each of these functors factors through \(C_{/U}\), we conclude that \(f^*\) itself must be conservative.Conversely, assume that \(f^*\) is conservative. We must show that the map \(\colim_n \check{C}_n(f) \to X\) is an isomorphism in \(C\). Since \(f^*\) is conservative and preserves groupoid colimits, it suffices to show that \(\colim_n \check{C}_n(f) \times_X U \to U\) is an isomorphism. This follows from the fact that the simplicial diagram \([n] \mapsto U^{\times^{(n+1)}_{/X}} \times_X U \iso U^{\times^{(n+2)}_{/X}}\) admits an extra degeneracy.(2) Consider a pullback square such that \(f\) is an effective epimorphism. Since the functor \(g^*\colon C_{/X} \to C_{/X'}\) preserves pullbacks, it sends \(\check{C}_{\bullet}(f)\) to \(\check{C}_{\bullet}(f')\). By assumption on \(C\), it also preserves groupoid colimits, hence we see that the map \(\colim_{n} \check{C}_n(f') \simeq (\colim_n \check{C}_n(f)) \times_X X' \to X \times_X X' = X'\) is an equivalence, implying that also \(f'\) is an effective epimorphism.
