Proof.
We will prove this by verifying (1) and (2') of Topologies, Lemma 34.10.12. The sheaf property for Zariski coverings follows from the fact that F has the sheaf property for all h coverings. Finally, suppose that X \to Y is a morphism of affine schemes over S such that \{ X \to Y\} is a V covering. By Lemma 61.10.2 we can write X = \mathop{\mathrm{lim}}\nolimits X_ i as a directed limit of affine schemes over Y such that \{ X_ i \to Y\} is an h covering for each i. We obtain
\begin{align*} & \text{Equalizer}( \xymatrix{ F(X) \ar@<1ex>[r] \ar@<-1ex>[r] & F(X \times _ Y X) } ) \\ & = \text{Equalizer}( \xymatrix{ \mathop{\mathrm{colim}}\nolimits F(X_ i) \ar@<1ex>[r] \ar@<-1ex>[r] & \mathop{\mathrm{colim}}\nolimits F(X_ i \times _ Y X_ i) } ) \\ & = \mathop{\mathrm{colim}}\nolimits \text{Equalizer}( \xymatrix{ F(X_ i) \ar@<1ex>[r] \ar@<-1ex>[r] & F(X_ i \times _ Y X_ i) } ) \\ & = \mathop{\mathrm{colim}}\nolimits F(Y) = F(Y) \end{align*}
which is what we wanted to show. The first equality because F is limit preserving and X = \mathop{\mathrm{lim}}\nolimits X_ i and X \times _ Y X = \mathop{\mathrm{lim}}\nolimits X_ i \times _ Y X_ i. The second equality because filtered colimits are exact. The third equality because F satisfies the sheaf property for h coverings.
\square
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