Disjoint coproducts and descent

Example 23.7.9 (Disjoint coproducts). For an \(\infty \)-category \(T\) with arbitrary coproducts, we say that coproducts are disjoint if for all \(X, Y \in T\) the diagram

is a pullback square. We claim that arbitrary coproducts in \(T\) satisfy descent if and only if they are universal and \(T\) has disjoint coproducts.

By Lemma 23.7.7, coproducts satisfy descent if and only if for every collection of maps \(\{Y_i \to X_i\}_{i \in I}\) the commutative square

is a pullback square for every \(j \in I\). To see that this reduces to disjointness, consider \(X' := \coprod _{i \in I \setminus \{j\}} X_i\) and \(Y' := \coprod _{i \in I \setminus \{j\}} Y_i\). We may write \(X = X_j \sqcup X'\) and \(Y = Y_j \sqcup Y'\). Invoking universality of coproducts once more shows that the condition reduces to disjointness of binary coproducts.