The Stacks project

59.13 Sheafification

There is a canonical map $\mathcal{F}(U) \to \check H^0 (\mathcal{U}, \mathcal{F})$, $s \mapsto (s |_{U_ i})_{i\in I}$. We say that a morphism of coverings from a covering $\mathcal{V} = \{ V_ j \to V\} _{j \in J}$ to $\mathcal{U}$ is a triple $(\chi , \alpha , \chi _ j)$, where $\chi : V \to U$ is a morphism, $\alpha : J \to I$ is a map of sets, and for all $j \in J$ the morphism $\chi _ j$ fits into a commutative diagram

Given the data $\chi , \alpha , \{ \chi _ j\} _{j \in J}$ we define

We then claim that

1. the map is well-defined, and

2. depends only on $\chi$ and is independent of the choice of $\alpha , \{ \chi _ j\} _{j \in J}$.

We omit the proof of the first fact. To see part (2), consider another triple $(\psi , \beta , \psi _ j)$ with $\chi = \psi$. Then we have the commutative diagram

Given a section $s \in \mathcal{F}(\mathcal{U})$, its image in $\mathcal{F}(V_ j)$ under the map given by $(\chi , \alpha , \{ \chi _ j\} _{j \in J})$ is $\chi _ j^*s_{\alpha (j)}$, and its image under the map given by $(\psi , \beta , \{ \psi _ j\} _{j \in J})$ is $\psi _ j^*s_{\beta (j)}$. These two are equal since by assumption $s \in \check H^0(\mathcal{U}, \mathcal{F})$ and hence both are equal to the pullback of the common value

pulled back by the map $(\chi _ j, \psi _ j)$ in the diagram.

In other words, this means that the natural map $\mathcal{F} \to \mathcal{F}^\#$ is a left adjoint to the forgetful functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \textit{PSh}(\mathcal{C})$.