General sheafification

In this posting, we follow Vistoli's notes to study a sheafification of a presheaf $F$ on a site $\mathcal{C}$ (i.e., a functor $F : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$). We would like to construct a sheaf $F^{\mathrm{a}} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ equipped with a map $a : F \rightarrow F^{\mathrm{a}}$ of functors such that for any sheaf $G$ and a map $\phi : F^{\mathrm{a}} \rightarrow G,$ there exists a unique map $\phi : F^{\mathrm{a}} \rightarrow G$ with a factorization $F \xrightarrow{a} F^{\mathrm{a}} \xrightarrow{\phi} G$ of $\phi$. It is immediate that a sheafification is unique up to an isomorphism if it ever exists.

Necessary conditions. Instead, we look for a sheaf $F^{\mathrm{a}}$ together with a map $a : F \rightarrow F^{\mathrm{a}}$ such that given any object $U$ of $\mathcal{C},$

1. if $\xi, \eta \in F(U)$ with $a_{U}(\xi) = a_{U}(\eta),$ then there exists a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $\xi|_{U_{i}} = \eta|_{U_{i}}$ for all $i \in I,$ and
2. if $\omega \in F^{a}(U),$ then there exists a cover $\{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $\omega|_{U_{i}} = a_{U_{i}}(\xi_{i})$ for all $i \in I.$

Remark. In a colloquial term, the sheafification of a presheaf remembers local data of each section of the presheaf.

Why are these necessary conditions? Given the condition above, let $\psi : F \rightarrow G$ any map where $G$ is a sheaf. Fix any object $U$ of $\mathcal{C}.$ Given $\omega \in F^{a}(U),$ we may choose a cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $\omega|_{U_{i}} = a(\xi_{i})$ for all $i \in I.$ We have $$\begin{align*}a_{U_{ij}}(\xi_{i}|_{U_{ij}}) &= a_{U_{i}}(\xi_{i})|_{U_{ij}} \\ &= \omega|_{U_{ij}} \\ &= a_{U_{j}}(\xi_{j})|_{U_{ij}} \\ &= a_{U_{ij}}(\xi_{j}|_{U_{ij}})\end{align*}$$ in $F^{a}(U_{ij}).$ Thus, there is a cover $\{V_{s} \rightarrow U_{ij}\}_{s \in J}$ such that $$\xi_{i}|_{V_{s}} = \xi_{j}|_{V_{s}}$$ for all $s \in J.$ This implies that $$\psi_{U_{i}}(\xi_{i})|_{V_{s}} = \psi_{V_{s}}(\xi_{i}|_{V_{s}}) = \psi_{V_{s}}(\xi_{j}|_{V_{s}}) = \psi_{U_{j}}(\xi_{j})|_{V_{s}}$$ for all $s \in J,$ so we must have $$\psi_{U_{i}}(\xi_{i})|_{U_{ij}} = \psi_{U_{j}}(\xi_{j})|_{U_{ij}}$$ in $G(U_{ij})$ for $i, j \in I.$ Since $G$ is a sheaf, this implies that there is a unique $\tilde{\psi}_{\mathscr{U}}(\omega) \in G(U)$ such that $\tilde{\psi}_{\mathscr{U}}(\omega)|_{U_{i}} = \psi_{U_{i}}(\xi_{i})$ for all $i \in I.$ If we have two covers of $U,$ we may take their refinements, and since $G$ is a sheaf, this implies that $\tilde{\psi}_{\mathscr{U}}(\omega)$ does not depend on the choice of the cover $\mathscr{U},$ so we may write $\tilde{\psi}_{U}(\omega) := \tilde{\psi}_{\mathscr{U}}(\omega)$ instead. Hence, we have defined a map $\tilde{\psi}_{U} : F^{a}(U) \rightarrow G(U).$ If $V \rightarrow U$ is a map in $\mathcal{C}$ and $s \in F^{a}(U),$ then taking a cover $\{U_{i} \rightarrow U\}_{i \in I}$ for $U$ to get a cover $\{V_{i} \rightarrow V\}_{i \in I}$ for $V$ where $V_{i} = V \times_{U} U_{i},$ we have $$(\tilde{\psi}_{U}(s)|_{V})|_{V_{i}} = \tilde{\psi}_{U}(s)|_{V_{i}} = \tilde{\psi}_{V_{i}}(s) = \tilde{\psi}_{V}(s)|_{V_{i}}$$ for all $i \in I,$ so $\tilde{\psi}_{U}(s)|_{V} = \tilde{\psi}_{V}(s).$ Thus, our definition gives a map $\tilde{\psi} : F^{a} \rightarrow G$ of sheaves. The definition of $\tilde{\psi}$ immediately tells us that $\tilde{\psi} \circ a = \psi.$ To show uniqueness, say we have another $\psi' : F^{a} \rightarrow G$ such that $\psi' \circ a = \psi.$ Then given any $\omega \in F^{a}(U),$ we choose a cover $\{U_{i} \rightarrow U\}_{i \in I}$ with elements $\xi_{i} \in F(U_{i})$ such that $a_{U_{i}}(\xi_{i}) = \omega|_{U_{i}}$ for $i \in I.$ This implies that $$\begin{align*}\psi'_{U}(\omega)|_{U_{i}} &= \psi'_{U_{i}}(\omega|_{U_{i}}) \\ &= \psi'_{U_{i}}(a_{U_{i}}(\xi_{i})) \\ &= \psi_{U_{i}}(\xi_{i}) \\ &= \tilde{\psi}_{U_{i}}(a_{U_{i}}(\xi_{i})) \\ &= \tilde{\psi}_{U_{i}}(\omega|_{U_{i}}) \\ &= \tilde{\psi}_{U}(\omega)|_{U_{i}}\end{align*}$$ for all $i \in I,$ so $\psi'_{U}(\omega) = \tilde{\psi}_{U}(\omega).$ Hence, we have $\psi' = \tilde{\psi},$ showing the uniqueness.

Sheafifications valued in different categories. The discussion above applies even when we replace $\textbf{Set}$ with $\textbf{Ab}$ or the category of modules over a ring.

Construction. For any $s, t \in F(U),$ we write $s \sim t$ if there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I.$ This is an equivalence relation where we can check transitivity by taking the joint cover of two given covers.

Define $F^{\mathrm{s}}(U)$ be the set of equivalence classes of elements in $F(U)$ with respect to this equivalence relation. Given any morphism $V \rightarrow U$ in $\mathcal{C},$ if $s \sim t$ in $F(U),$ we have a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I,$ so $(s|_{V})|_{V_{i}} = (t|_{V})|_{V_{i}}$ for all $i \in I,$ where $\{V \times_{U} U_{i} = V_{i} \rightarrow V\}_{i \in I}$ is a cover for $V.$ Hence, we have $s|_{V} \sim t|_{V}$ in $F(V).$ This lets us define $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ by $[s] \mapsto [s|_{V}].$ This immediately makes $F^{\mathrm{s}} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ a functor (i.e., a presheaf). If $\{U_{i} \rightarrow U\}_{i \in I}$ is a cover and $s, t \in F(U)$ with $s|_{U_{i}} \sim t|_{U_{i}} \in F(U_{i})$ for all $i \in I,$ then we may take a cover $\{V_{ij} \rightarrow U_{i}\}_{j \in J_{i}}$ for each $i \in I$ such that $s|_{V_{ij}} = t|_{V_{ij}}$ for all $j \in J_{i}.$ Since $\{V_{ij} \rightarrow U_{i} \rightarrow U\}_{i \in I, j \in J_{i}}$ is a cover for $U,$ we have $s \sim t$ in $F(U).$ This implies that $F^{\mathrm{s}}$ is a separated presheaf.

Given an object $U$ of $\mathcal{C},$ we consider the set of pairs $(\mathscr{U}, \boldsymbol{s}),$ where $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ is a cover of $U$ and $\boldsymbol{s} = ([s_{i}])_{i \in I} \in \prod_{i \in I}F^{\mathrm{s}}(U_{i})$ such that $[s_{i}|_{U_{ij}}] = [s_{j}|_{U_{ij}}] \in F^{\mathrm{s}}(U_{ij})$ for all $i, j \in I$ where $U_{ij} = U_{i} \times_{U} U_{j}.$ For two pairs $(\mathscr{U}, \boldsymbol{s})$ and $(\mathscr{V}, \boldsymbol{t}),$ we write $$(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{V}, \boldsymbol{t})$$ if $$[s_{i}|_{U_{i} \times_{U} V_{j}}] = [t_{j}|_{U_{i} \times_{U} V_{j}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} V_{j})$ for all $i \in I$ and $j \in J,$ where we wrote 

• $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I},$
• $\mathscr{V} = \{V_{j} \rightarrow U\}_{j \in J},$
• $\boldsymbol{s} = ([s_{i}])_{i \in I},$
• $\boldsymbol{t} = ([t_{j}])_{j \in J}.$

We claim that this gives an equivalence relation among such pairs. Reflexivity is evident, and for symmetry, we just need to use the isomorphisms $V_{j} \times_{U} U_{i} \simeq U_{i} \times_{U} V_{j}$ because the covers always include isomorphisms and we can compose covers to get another. For transitivity, suppose that $(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{V}, \boldsymbol{t})$ and $(\mathscr{V}, \boldsymbol{t}) \sim (\mathscr{W}, \boldsymbol{u}),$ where we write  

• $\mathscr{W} = \{W_{k} \rightarrow U\}_{k \in K},$
• $\boldsymbol{u} = ([u_{k}])_{k \in K}.$

We have $$[s_{i}|_{U_{i} \times_{U} V_{j}}] = [t_{j}|_{U_{i} \times_{U} V_{j}}]$$ for all $i \in I$ and $j \in J,$ while $$[t_{j}|_{V_{j} \times_{U} W_{k}}] = [u_{k}|_{V_{j} \times_{U} W_{k}}]$$ for all $j \in J$ and $k \in K.$ This gives us $$[s_{i}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [t_{j}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [u_{k}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}].$$ Running over $j \in V,$ since $F^{\mathrm{s}}$ is a separated presheaf, we necessarily have $$[s_{i}|_{U_{i} \times_{U} W_{k}}] = [u_{k}|_{U_{i} \times_{U} W_{k}}].$$ This shows the transitivity, so we indeed get an equivalence relation.

We define $F^{\mathrm{a}}(U)$ to be the set of equivalence classes $[(\mathscr{U}, \boldsymbol{s})]$ of the pairs we have discussed above. Given a morphism $V \rightarrow U$ in $\mathcal{C},$ we define $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ by $[(\mathscr{U}, \boldsymbol{s})] \mapsto [(V \times_{U} \mathscr{U}, \boldsymbol{s}^{*})],$ where $V \times_{U} \mathscr{U} = \{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ and $\boldsymbol{s}^{*} = (s_{i}|_{V \times U_{i}})_{i \in I}.$ Given $(\mathscr{U}, \boldsymbol{s}) \sim (\mathscr{W}, \boldsymbol{t}),$ we have $$[s_{i}|_{U_{i} \times_{U} W_{j}}] = [t_{j}|_{U_{i} \times_{U} W_{j}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} W_{j}),$ for all $i \in I$ and $j \in J.$ Since $$(V \times_{U} U_{i}) \times_{V} (V \times_{U} W_{j}) \simeq V \times_{U} U_{i} \times_{U} W_{j},$$ having $$[s_{i}|_{V \times_{U} U_{i} \times_{U} W_{j}}] = [t_{j}|_{V \times_{U} U_{i} \times_{U} W_{j}}]$$ is enough to check that the restrictions $F^{\mathrm{a}}(U) \rightarrow F^{\mathrm{a}}(V)$ are well-defined.

Given a cover $\{U_{i} \rightarrow U\}_{i \in I},$ fix any $[(\mathscr{V}, \boldsymbol{s})], [(\mathscr{W}, \boldsymbol{t})] \in F^{a}(U)$ such that $$[(U_{i} \times_{U} \mathscr{V}, \boldsymbol{s}^{*})] = [(U_{i} \times_{U}\mathscr{W}, \boldsymbol{t}^{*})]$$ for all $i \in I.$ Since $$(U_{i} \times_{U} V_{j}) \times_{U_{i}} (U_{i} \times_{U} W_{k}) \simeq U_{i} \times_{U} V_{j} \times_{U} W_{k},$$ we necessarily have $$[s_{j}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}] = [t_{k}|_{U_{i} \times_{U} V_{j} \times_{U} W_{k}}]$$ in $F^{\mathrm{s}}(U_{i} \times_{U} V_{j} \times_{U} W_{k})$. Ranging over all $i \in I,$ using the fact that $F^{\mathrm{s}}$ is a separated presheaf, we have $$[s_{j}|_{V_{j} \times_{U} W_{k}}] = [t_{k}|_{V_{j} \times_{U} W_{k}}]$$ in $F^{\mathrm{s}}(V_{j} \times_{U} W_{k})$. This implies that $$[(\mathscr{V}, \boldsymbol{s})] = [(\mathscr{W}, \boldsymbol{t})].$$ This shows that $F^{\mathrm{a}}$ is a separated presheaf.

To show that $F^{\mathrm{a}}$ is a sheaf, fix $[(\mathscr{V}_{i}, \boldsymbol{s}_{i})] \in F^{\mathrm{a}}(U_{i})$ for $i \in I$ such that $$[(U_{i} \times_{U} U_{j} \times_{U} \mathscr{V}_{i} , \boldsymbol{s}_{i}^{*})] = [(U_{i} \times_{U} U_{j} \times_{U} \mathscr{V}_{j}, \boldsymbol{s}_{j}^{*})].$$ We have $$(U_{i} \times_{U} U_{j} \times_{U} V_{i,l}) \times_{U_{i} \times_{U} U_{j}} (U_{i} \times_{U} U_{j} \times_{U} V_{j,m}) \simeq U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm},$$ so the above identity implies that we have $$s_{i,l} |_{U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm}} = s_{j,m} |_{U_{i} \times_{U} U_{j} \times_{U} V_{il} \times_{U} V_{jm}},$$ but then this makes $[(\mathscr{V}, \boldsymbol{s})] \in F^{\mathrm{a}}(U)$ given by $\mathscr{V} = \{V_{il} \rightarrow U_{i} \rightarrow U\}_{i,l}$ and $\boldsymbol{s} = ([s_{i,l}])_{i,l} \in \prod_{i,l} F^{\mathrm{s}}(V_{il}).$ This shows that $F^{\mathrm{a}}$ is a sheaf.

We define $a_{U} : F(U) \rightarrow F^{\mathrm{a}}(U)$ by $s \mapsto [(\{\mathrm{id}_{U}\}, (s))].$ It is immediate that this gives rise to a map $a : F \rightarrow F^{\mathrm{a}}$ of functors. Given $s, t \in F(U)$ with $$[(\{\mathrm{id}_{U}\}, (s))] = [(\{\mathrm{id}_{U}\}, (t))],$$ we have $[s] = [t] \in F^{\mathrm{s}}(U),$ so by our definition, there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = t|_{U_{i}}$ for all $i \in I.$ Any given $[(\mathscr{U}, \textbf{s})] \in F^{\mathrm{a}}(U)$ comes with the cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ and $\boldsymbol{s} = ([s_{i}])_{i \in I} \in \prod_{i \in I}F^{\mathrm{s}}(U_{i})$ so that $$a_{U_{i}}(s_{i}) = [(\{\mathrm{id}_{U_{i}}\}, (s_{i}))] = [(U_{i} \times_{U} \mathscr{U}, \boldsymbol{s}^{*})]| = [(\mathscr{U}, \boldsymbol{s})]|_{U_{i}},$$ where we get the second identity because $$U_{i} \times_{U_{i}} (U_{i} \times_{U} U_{j}) \simeq U_{i} \times_{U} U_{j}$$ and $s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}}$ for all $j \in J.$ Thus, we are done checking the necessary conditions for $a : F \rightarrow F^{\mathrm{a}}$ to be a sheafification of $F.$