## 59.13 Sheafification

Definition 59.13.1. Let $\mathcal{F}$ be a presheaf on the site $\mathcal{C}$ and $\mathcal{U} = \{ U_ i \to U\} \in \text{Cov} (\mathcal{C})$. We define the zeroth Čech cohomology group of $\mathcal{F}$ with respect to $\mathcal{U}$ by

$\check H^0 (\mathcal{U}, \mathcal{F}) = \left\{ (s_ i)_{i\in I} \in \prod \nolimits _{i\in I }\mathcal{F}(U_ i) \text{ such that } s_ i|_{U_ i \times _ U U_ j} = s_ j |_{U_ i \times _ U U_ j} \right\} .$

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

$\xymatrix{ V_ j \ar[rr]_{\chi _ j} \ar[d] & & U_{\alpha (j)} \ar[d] \\ V \ar[rr]^\chi & & U. }$

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

\begin{eqnarray*} \check H^0(\mathcal{U}, \mathcal{F}) & \longrightarrow & \check H^0(\mathcal{V}, \mathcal{F}) \\ (s_ i)_{i\in I} & \longmapsto & \left(\chi _ j^*\left(s_{\alpha (j)}\right)\right)_{j\in J}. \end{eqnarray*}

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

$\xymatrix{ V_ j \ar[rrr]_{(\chi _ j, \psi _ j)} \ar[dd] & & & U_{\alpha (j)} \times _ U U_{\beta (j)} \ar[dl] \ar[dr] \\ & & U_{\alpha (j)} \ar[dr] & & U_{\beta (j)} \ar[dl] \\ V \ar[rrr]^{\chi = \psi } & & & U. }$

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

$s_{\alpha (j)}|_{U_{\alpha (j)} \times _ U U_{\beta (j)}} = s_{\beta (j)}|_{U_{\alpha (j)} \times _ U U_{\beta (j)}}$

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

Theorem 59.13.2. Let $\mathcal{C}$ be a site and $\mathcal{F}$ a presheaf on $\mathcal{C}$.

1. The rule

$U \mapsto \mathcal{F}^+(U) := \mathop{\mathrm{colim}}\nolimits _{\mathcal{U} \text{ covering of }U} \check H^0(\mathcal{U}, \mathcal{F})$

is a presheaf. And the colimit is a directed one.

2. There is a canonical map of presheaves $\mathcal{F} \to \mathcal{F}^+$.

3. If $\mathcal{F}$ is a separated presheaf then $\mathcal{F}^+$ is a sheaf and the map in (2) is injective.

4. $\mathcal{F}^+$ is a separated presheaf.

5. $\mathcal{F}^\# = (\mathcal{F}^+)^+$ is a sheaf, and the canonical map induces a functorial isomorphism

$\mathop{\mathrm{Hom}}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{F}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}^\# , \mathcal{G})$

for any $\mathcal{G} \in \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$.

Proof. See Sites, Theorem 7.10.10. $\square$

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

## Comments (7)

Comment #236 by Keenan Kidwell on

In 03NR, "where $\chi:\mathcal{V}\rightarrow\mathcal{U}$ is a morphism" should be "where $\chi:V\rightarrow U$ is a morphism."

Comment #2810 by Dario Weißmann on

Typo(s) in the paragraph after Definition 53.13.1:

• "Given the data $\chi, \alpha, \{\chi_j}_{i\in J}\}$..." should read "...$\{\chi_j}_{j\in J}\}$

• same thing happens in (2)

• "...under the map given by $(\chi,\alpha,\{\chi_j}_{i\in J}\})$"

• "...under the map given by $(\psi,\beta,\{\psi_j}_{i\in J}\})$"

Comment #2811 by Dario Weißmann on

Typo(s) in the paragraph after Definition 53.13.1:

• "Given the data $\chi, \alpha, \{\chi_j\}_{i\in J}$..." should read "...$\{\chi_j\}_{j\in J}$

• same thing happens in (2)

• "...under the map given by $(\chi,\alpha,\{\chi_j\}_{i\in J})$"

• "...under the map given by $(\psi,\beta,\{\psi_j\}_{i\in J})$"

Oh, and the preview function doesn't work (just for me?).

Comment #2912 by on

The preview also stopped working for me... This could be because of our change to https? Pieter?

Typos fixed here.

Comment #6614 by Liu S-H on

"These two are equal since by assumption $s \in \check H(\mathcal{U}, \mathcal{F})$ and" should be "$s \in \check H^0(\mathcal{U}, \mathcal{F})$"?

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