A Note On Sheafification

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A presheaf ${\mathcal{F}}$ on a topological space ${X}$ with values in a category ${\mathcal{D}}$ (usually with values in GrpCommRing, or Set), can be associated to a sheaf ${\widetilde{\mathcal{F}}}$ on ${X}$ in a natural way. In fact, this association is left adjoint to the inclusion functor from the category of sheaves on ${X}$ to the category of presheaves on ${X}$.

The standard construction of ${\widetilde{\mathcal{F}}}$ is fairly straightforward: consider the topology on the disjoint union of stalks ${\text{Tot}(\mathcal{F}) = \bigcup_{x \in X} \mathcal{F}_x}$ generated by the collection of sets

$\displaystyle \begin{array}{rcl} \mathfrak{V}(b, U) &=& \{ (b_x, x) : x \in U\}, \end{array}$

where ${U \subseteq X}$ is open, ${b\in \mathcal{F}(U)}$ and ${b_x}$ is the image of ${b}$ in ${\mathcal{F}_x}$. Let ${p : \text{Tot}(\mathcal{F}) \rightarrow X}$ be the natural projection. For every ${U \subseteq X}$ define ${\widetilde{\mathcal{F}}(U)}$ to be the set of continuous sections ${s : U \rightarrow \text{Tot}(\mathcal{F})}$ of ${p}$, i.e. ${p\circ s = \text{id}_U}$. It follows that ${\widetilde{\mathcal{F}}}$ is a sheaf and there is a natural morphism ${i : \mathcal{F} \rightarrow \widetilde{\mathcal{F}}}$, such that ${\mathcal{F}}$ is a sheaf if, and only if, ${i}$ is an isomorphism, (this is a fairly easy exercise).

This shows that every sheaf arises as the sheaf of sections of some continuous map onto ${X}$. This also justifies the terminology: an element ${a \in \mathcal{F}(U)}$ is called a section. At first glance, it is hard to see how this sheaf relates to ${\mathcal{F}}$. Indeed, the canonical map ${i}$ fails to be either surjective or injective, in general. This says that we cannot “complete” ${\mathcal{F}}$ by simply adding more elements, or by glueing together elements; we must do both. Thus, we can can’t view ${\mathcal{F}}$ as a subpresheaf of ${\widetilde{\mathcal{F}}}$ or as a cover. However, there is another natural way to think of ${\widetilde{\mathcal{F}}}$.

Let ${\mathcal{G}}$ be a sheaf on ${X}$ with values in some category ${\mathcal{D}}$, as above. One of the advantages of having a sheaf on ${X}$ is that a section ${s \in \mathcal{G}(U)}$ is uniquely determined by the collection of elements ${\{s_x \in \mathcal{G}_x : x \in U\}}$. This can fail for presheaves. Thus, we need ${i(U): \mathcal{F}(U) \rightarrow \widetilde{\mathcal{F}}(U)}$ to glue together elements ${a, b \in \mathcal{F}(U)}$ such that ${a_x = b_x}$ for all ${x \in U}$. This is evident in the commutative diagram:

$\displaystyle \begin{array}{rcl} \begin{array}{ccc} \mathcal{F}(U) & \longrightarrow & \widetilde{\mathcal{F}}(U) \\ \downarrow & \; & \downarrow \\ \mathcal{F}_x & \longrightarrow & \widetilde{\mathcal{F}}_x \end{array} \end{array}$

This says that ${i(U)(a)}$ and ${i(U)(b)}$ have the same image in ${\widetilde{\mathcal{F}}_x}$ for all ${x \in U}$. Thus, because ${\widetilde{\mathcal{F}}}$ is a sheaf, ${i(U)(a) = i(U)(b)}$.

Aside from glueing sections together that “should” be equal, we also need to add more elements to our presheaf ${\mathcal{F}}$. Suppose we have an open cover of an open subset ${U \subseteq X}$: ${U = \bigcup_\alpha U_\alpha}$. If we can find a collection ${\{b_\alpha\}, b_\alpha \in U_\alpha}$, such that ${b_\alpha|_{U_\alpha \cap U_\beta} = b_{\beta}|_{U_\alpha \cap U_\beta}}$, then we should be able to find a “global” section ${b}$ on ${U}$ such that ${b|_{U_\alpha} = b_\alpha}$. Because our presheaf may be too “small” and may not contain this element, we need to ensure that our sheafification ${\widetilde{\mathcal{F}}}$ does. The set ${\{b_\alpha\}}$ is compatible in the obvious sense, so we can glue them together to form a section ${\hat{b} \in \widetilde{\mathcal{F}}(U), \hat{b} : x \mapsto (b_{\alpha})_x}$. This map is continuous because ${\hat{b}^{-1}(\mathfrak{V}(a, W)) = \{x \in W \cap U : \hat{b}(x) = a_x\}}$. is open. Indeed, if ${a_x = b(x) = (b_\alpha)_x}$ for some ${x}$, then there is an open subsets ${W' \subseteq W \cap U}$ such that ${x\in W'}$ and ${a|_{W'} = (b_\alpha)|_{W'}}$ in ${\mathcal{F}(W')}$, so ${W' \subseteq \hat{b}^{-1}(\mathfrak{V}(a, W))}$. It is also clear that ${p \circ \hat{b} = \text{id}_U}$.

To sum up our construction, we note the following: to construct ${\widetilde{\mathcal{F}}}$ we find the most “efficient” sheaf such that

1. sections that have the same germ at every point are equal
2. compatible sets of sections have a common extension.

This gives the following alternative description of ${\widetilde{\mathcal{F}}}$:

$\displaystyle \begin{array}{rcl} \widetilde{\mathcal{F}}(U) &=& \{ \{(b_{\alpha}, W_\alpha)\}_{\alpha} : U = \bigcup_{\alpha} W_\alpha, b_\alpha \in \mathcal{F}(W_{\alpha}), (b_\alpha)_x = (b_\beta)_x \text{ for all } x \in W_\alpha \cap W_\beta\} /\sim \end{array}$

where ${\{(b_\alpha, W_\alpha)\}_\alpha \sim \{(a_\beta, V_{\beta})\}_\beta}$, if for all ${\alpha, \beta}$ and ${x \in W_\alpha \cap V_{\beta}}$, ${(b_\alpha)_x = (a_\beta)_x}$. Note that condition ${2}$ and the assumption that this extension is unique is just the sheaf axiom and, hence, it implies ${1}$.

We claim that this definition of ${\widetilde{\mathcal{F}}}$ is equivalent to the definition given above. Indeed, we can view each collection ${\{(b_\alpha, W_\alpha)\}_\alpha}$ as a section ${\hat{b}}$ in the natural way: ${\hat{b}(x) = (b_\alpha)_x}$ if ${x \in W_\alpha}$. This is independent of the representative of ${\{(b_\alpha, W_\alpha)\}_\alpha}$ by ${\sim}$. Note that ${\hat{b}(x)}$ is continuous, as we showed earlier. Further, any section ${s : U \rightarrow \text{Tot}(\mathcal{F})}$ arises in this way. Indeed, let ${x \in U}$, then ${s(x) \in \mathfrak{V}(b^x, W_x)}$ for a neighborhood ${W_x}$ of ${x}$ and an element ${b^x \in \mathcal{F}(W_x)}$. Thus, because ${s}$ is continuous there exists ${W_x' \subseteq s^{-1}( \mathfrak{V}(b^x, W_x)) = \{ y \in W \cap U : s(y) = (b^x)_y\}}$. Consider the collection ${b = \{(b^x, W_x')\}_{x\in U}}$. Because ${(b^x)_y = s(y) = (b^z)_y}$ for all ${y \in W_x' \cap W_z'}$, it follows that ${b}$ corresponds to ${s}$. This shows that the correspondence is surjective. We claim that this correspondence is also injective. Indeed, if ${\{(b_\alpha, W_\alpha)\}_\alpha}$ and ${\{(a_\beta, V_\beta)\}_\beta}$ correspond to the same section ${s}$, then ${(b_\alpha)_x = s(x) = (a_\beta)_x}$ for all ${x \in W_\alpha \cap V_\beta}$. Thus, ${\{(b_\alpha, W_\alpha)\}_\alpha = \{(a_\beta, V_\beta)\}_\beta}$ Finally we note that this correspondence is actually a morphism of presheaves: Restricting a collection ${\{(b_\alpha, W_\alpha)\}_\alpha \in \widetilde{\mathcal{F}}(U)}$ to a subset ${U'\subseteq U}$, yields the element ${\{(b_\alpha|_{U'}, W_\alpha\cap U')\}_{\alpha}}$ and this corresponds to the restricted section ${\hat{b}|_{U'}}$.

This new definition has some utility. It helps to illuminate the following claims:

Claim 1 Let ${i : \mathcal{F} \rightarrow \widetilde{\mathcal{F}}}$ be the natural map: ${i(U)(b) = \{(b, U)\}}$. Then ${\mathcal{F}}$ is a sheaf if, and only if, ${i}$ is an isomorphism.

Proof: If ${\mathcal{F}}$ is a sheaf then each collection ${\{(b_{\alpha}, W_\alpha)\}_{\alpha} \in \widetilde{\mathcal{F}}(U)}$ can be written uniquely as ${\{(b, U)\}}$ for some ${b \in \mathcal{F}(U)}$. Conversely, suppose ${i}$ is an isomorphism, let ${U \subseteq X}$ be an open subset, let ${U = \bigcup W_\alpha}$, and let ${\{b_\alpha : b_\alpha \in \mathcal{F}(W_\alpha)\}}$ be a collection such that ${b_{\alpha}|_{W_\alpha \cap W_\beta} = b_{\beta}|_{W_\alpha \cap W_\beta}}$. Then ${(b_{\alpha})_x = (b_\beta)_x}$ for all ${x \in W_\alpha \cap W_\beta}$, so ${\{(b_\alpha, W_\alpha)\}_\alpha \in \widetilde{\mathcal{F}}(U)}$. Thus, there exists a unique element ${b \in \mathcal{F}(U)}$ such that ${i(U)(b)= \{(b_\alpha, W_\alpha)\}_\alpha}$. Therefore, for each ${\alpha}$, ${b_x = (b_\alpha)_x}$ for all ${x \in W_\alpha}$. In particular, for each fixed ${\alpha}$, ${i(W_\alpha)(b|_{W_\alpha}) = \{(b|_{W_\alpha}, W_\alpha)\} = \{(b_\alpha, W_\alpha)\} = i(W_\alpha)(b_\alpha)}$, whence, ${b|_{W_\alpha} = b_\alpha}$, as desired. $\Box$

Now, suppose that ${\mathcal{G}}$ is a sheaf on ${X}$. Let ${j : \mathcal{F} \rightarrow \mathcal{G}}$ be a morphism of presheaves.

Claim 2 There exists a unique map ${\widetilde{j} : \widetilde{\mathcal{F}} \rightarrow \mathcal{G}}$, such that ${\widetilde{j} \circ i = j}$.

Proof: Let ${U \subseteq X}$ be an open subset and let ${\{(b_\alpha, W_\alpha)\}_\alpha \in \widetilde{\mathcal{F}}(U)}$. Observe that the commutative diagrams

$\displaystyle \begin{array}{rcl} \begin{array}{ccc} \mathcal{F}(W_\alpha) & \stackrel{j(W_\alpha)}{\longrightarrow} & \mathcal{G}(W_\alpha) \\ \downarrow & \; & \downarrow \\ \mathcal{F}(W_\alpha\cap W_\beta) & \stackrel{j(W_\alpha\cap W_\beta)}{\longrightarrow} & \mathcal{G}(W_\alpha\cap W_\beta) \\ \downarrow & \; & \downarrow \\ \mathcal{F}_x & \longrightarrow & \mathcal{G}_x \end{array} \end{array}$

ensure that ${j(W_\alpha)(b_\alpha)_x = j(W_\beta)(b_\beta)_x}$ for all ${x \in W_\alpha \cap W_\beta}$. Thus, ${j(W_\alpha)(b_\alpha)|_{W_\alpha \cap W_\beta} = j(W_\beta)(b_\beta)|_{W_\alpha \cap W_\beta}}$ and so by the sheaf axiom for ${\mathcal{G}}$, there is a unique extension ${\widetilde{j}(U)(\{(b_\alpha, W_\alpha)\}_\alpha) \in \mathcal{G}(U)}$. For convenience, we can view ${\mathcal{G}}$ as ${\widetilde{\mathcal{G}}}$, with ${\widetilde{j}(U)(\{(b_\alpha, W_\alpha)\}_\alpha) =\{( j(U)(b_\alpha), W_\alpha)\}_\alpha}$. Now, if ${U' \subseteq U}$ is an open set, then

$\displaystyle \begin{array}{rcl} \widetilde{j}(U')( \{(b_\alpha, W_\alpha)\}_\alpha|_{U'}) &=& \widetilde{j}(U')(\{(b_\alpha|_{U'}, W_\alpha\cap U')\}_\alpha) \\ &=& \{(j(U')(b_\alpha|_{U'}), W_\alpha\cap U')\}_\alpha \\ &=& \{(j(U)(b_\alpha)|_{U'}, W_\alpha\cap U')\}_\alpha \\ &=& \{(j(U)(b_\alpha), W_\alpha )\}_\alpha|_{U'} \\ &=& \widetilde{j}(U) (\{(b_\alpha, W_\alpha)\})|_{U'}. \end{array}$

Thus, ${\widetilde{j}}$ is a morphism ${\widetilde{j} : \widetilde{\mathcal{F}} \rightarrow \mathcal{G}}$. From the definition of ${\widetilde{j}}$, it is clear that ${\widetilde{j} \circ i = j}$ and ${\widetilde{j}}$ is unique. $\Box$