Let \(C\) be a category which has finite limits, and let \(T\) be a presentable category satisfying the Giraud axioms. Let \(f\colon C \to T\) be a left exact functor. Then the unique colimit-preserving extension \(F\colon \PSh(C) \to T\) of \(f\) is left exact.
Proof
We say that a presheaf \(X \in \PSh(C)\) is good if the functor \(F\) preserves every pullback square of the form By universality of colimits, \(X\) is good if and only if \(F\) preserves all such pullback squares where \(Y\) and \(Z\) are representable. In particular, since \(f\) preserves pullbacks, every representable presheaf is good. Since coproducts are universal and disjoint in \(T\), they are van Kampen by Example 2.12; it follows that coproducts of good objects are good.Every presheaf admits an effective epimorphism from a coproduct of representables. Indeed, one may choose a representative of every connected component of every anima \(X(c)\), and the resulting map
\[\coprod_{(c,x)} h_c \longrightarrow X\]
is pointwise surjective on \(\pi_0\). It therefore remains to prove the following descent statement: if \(X_0 \twoheadrightarrow X\) is an effective epimorphism and \(X_0\) is good, then \(X\) is good.Let \(X_{\bullet} \to X\) be the augmented Čech nerve of \(p\colon X_0 \to X\). The pullback identities expressing that \(X_{\bullet}\) is a groupoid object are pullback squares over \(X_0\), so goodness of \(X_0\) implies that \(F(X_{\bullet})\) is a groupoid object. Since \(F\) preserves colimits, \(F(X_{\bullet}) \to F(X)\) is a colimit diagram. Effectivity of groupoids in \(T\) therefore identifies it with the augmented Čech nerve of \(F(p)\colon F(X_0) \to F(X)\). In particular, \(F(p)\) is an effective epimorphism.We record one consequence. Suppose that a map \(W \to X\) admits a lift \(W \to X_0\). Using either projection \(X_0\times_XX_0\to X_0\) as appropriate, there is a pullback square over \(X_0\) identifying
To test that \(X\) is good, we may take \(Y\) and \(Z\) to be representable. Since \(p\) is pointwise surjective on \(\pi_0\), the maps \(Y \to X\) and \(Z \to X\) admit lifts through \(X_0\) in the corresponding mapping animae. The same is then true of \(Y\times_XZ\to X\). Applying Equation (1) to these three objects, and using goodness of \(X_0\) for the pullback square
becomes an isomorphism after pullback along \(F(X_0)\to F(X)\). This map is an effective epimorphism, so the corresponding pullback functor is conservative. The comparison map is therefore itself an isomorphism. Thus \(X\) is good, completing the proof.
