Sifted colimits commute with finite products

Proof. Let \(I\) be a sifted \(\infty \)-category and let \(X,Y\colon I \to C\) be two functors. We need to show that the map \[ \colim _{i \in I} (X_i \times Y_i) \to (\colim _{i \in I} X_i) \times (\colim _{i \in I} Y_i) \] is an equivalence. This follows from the following computation: \begin {align*} \colim _{i \in I} (X_i \times Y_i) &\overset {(1)}{\simeq } \colim _{(i,j) \in I \times I} (X_i \times Y_j) \\ &\overset {(2)}{\simeq }\colim _{i \in I} \colim _{j \in I} (X_i \times Y_j) \\ &\overset {(3)}{\simeq } \colim _{i \in I} (X_i \times \colim _{j \in I} Y_j) \\ &\overset {(4)}{\simeq } (\colim _{i \in I} X_i) \times (\colim _{j \in I} Y_j). \end {align*}

Here equivalence (1) holds because the diagonal functor \(I \to I \times I, i \mapsto (i,i)\) is final by assumption on \(I\). The equivalence (2) is an instance of Lemma 21.2.10. The equivalences (3) and (4) hold because the product in \(C\) preserves colimits in each variable. □