Lemma 7.50.3. Assumption and notation as in Theorem 7.50.2. Then $J \subset J'$ if and only if every sheaf for the topology $J'$ is a sheaf for the topology $J$.

**Proof.**
One direction is clear. For the other direction suppose that $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J') \subset \mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J)$. By formal nonsense this implies that if $\mathcal{F}$ is a presheaf of sets, and $\mathcal{F} \to \mathcal{F}^\# $, resp. $\mathcal{F} \to \mathcal{F}^{\# , \prime }$ is the sheafification wrt $J$, resp. $J'$ then there is a canonical map $\mathcal{F}^\# \to \mathcal{F}^{\# , \prime }$ such that $\mathcal{F} \to \mathcal{F}^\# \to \mathcal{F}^{\# , \prime }$ equals the canonical map $\mathcal{F} \to \mathcal{F}^{\# , \prime }$. Of course, $\mathcal{F}^\# \to \mathcal{F}^{\# , \prime }$ identifies the second sheaf as the sheafification of the first with respect to the topology $J'$. Apply this to the map $S \to h_ U$ of Lemma 7.50.1. We get a commutative diagram

And clearly, if $S$ is a covering sieve for the topology $J$ then the middle vertical map is an isomorphism (by the lemma) and we conclude that the right vertical map is an isomorphism as it is the sheafification of the one in the middle wrt $J'$. By the lemma again we conclude that $S$ is a covering sieve for $J'$ as well. $\square$

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