Lemma 7.20.2. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be cocontinuous. Let $\mathcal{F}$ be a sheaf on $\mathcal{C}$. Then ${}_ pu\mathcal{F}$ is a sheaf on $\mathcal{D}$, which we will denote ${}_ su\mathcal{F}$.

**Proof.**
Let $\{ V_ j \to V\} _{j \in J}$ be a covering of the site $\mathcal{D}$. We have to show that

is an equalizer diagram. Since ${}_ pu$ is right adjoint to $u^ p$ we have

Hence it suffices to show that

becomes a coequalizer diagram after sheafification. (Recall that a coproduct in the category of sheaves is the sheafification of the coproduct in the category of presheaves, see Lemma 7.10.13.)

We first show that the second arrow of (7.20.2.1) becomes surjective after sheafification. To do this we use Lemma 7.11.2. Thus it suffices to show a section $s$ of $u^ ph_ V$ over $U$ lifts to a section of $\coprod u^ p h_{V_ j}$ on the members of a covering of $U$. Note that $s$ is a morphism $s : u(U) \to V$. Then $\{ V_ j \times _{V, s} u(U) \to u(U)\} $ is a covering of $\mathcal{D}$. Hence, as $u$ is cocontinuous, there is a covering $\{ U_ i \to U\} $ such that $\{ u(U_ i) \to u(U)\} $ refines $\{ V_ j \times _{V, s} u(U) \to u(U)\} $. This means that each restriction $s|_{U_ i} : u(U_ i) \to V$ factors through a morphism $s_ i : u(U_ i) \to V_ j$ for some $j$, i.e., $s|_{U_ i}$ is in the image of $u^ ph_{V_ j}(U_ i) \to u^ ph_ V(U_ i)$ as desired.

Let $s, s' \in (\coprod u^ ph_{V_ j})^\# (U)$ map to the same element of $(u^ ph_ V)^\# (U)$. To finish the proof of the lemma we show that after replacing $U$ by the members of a covering that $s, s'$ are the image of the same section of $\coprod u^ p h_{V_ j \times _ V V_{j'}}$ by the two maps of (7.20.2.1). We may first replace $U$ by the members of a covering and assume that $s \in u^ ph_{V_ j}(U)$ and $s' \in u^ ph_{V_{j'}}(U)$. A second such replacement guarantees that $s$ and $s'$ have the same image in $u^ ph_ V(U)$ instead of in the sheafification. Hence $s : u(U) \to V_ j$ and $s' : u(U) \to V_{j'}$ are morphisms of $\mathcal{D}$ such that

is commutative. Thus we obtain $t = (s, s') : u(U) \to V_ j \times _ V V_{j'}$, i.e., a section $t \in u^ ph_{V_ j \times _ V V_{j'}}(U)$ which maps to $s, s'$ as desired. $\square$

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