# When can stalks be glued to recover a sheaf?

Let $\mathcal{F}$ be a sheaf over some topological space. The stalks are $\mathcal{F}_x= \underset{{x\in U}}{ \underrightarrow{\lim}} \mathcal{F}(U)$. Is there a special name for a sheaf that satisfies $\mathcal{F}(U) = \underset{{x\in U}}{ \underleftarrow{\lim}} \mathcal{F}_x$?

Obviously this is a very restrictive property but here's a possible example:

Let $X=Spec A$ be an affine integral scheme with structure sheaf $\mathcal{O}_X$. We have:

$$\mathcal{O}_{X,x}= \underset{{x\in X_f}}{ \underrightarrow{\lim}} \mathcal{O}_X(X_f)=\underset{{f \notin \mathfrak{p}_x}}{\underrightarrow{\lim}} A_f = \bigcup_{f \notin \mathfrak{p}_x} A_f$$ But we also have (I hope):

$$\mathcal{O}_X(X_f)=A_f= \bigcap_{f \notin \mathfrak{p}_x \subset A} A_{\mathfrak{p}_x}=\underset{{f \notin \mathfrak{p}_x}}{\underleftarrow{\lim}} A_{\mathfrak{p}_x}=\underset{{x \in X_f}}{\underleftarrow{\lim}} \mathcal{O}_{X,x}$$

So we can recover the structure sheaf as a limit of the stalks. Does it still hold for non affine scheme? More generally:

When is a sheaf the inverse limit of its stalks?

Can I turn this into a technique for constructing sheaves?

Let $F: |X| \to Ab$ be a functor from the category of points of $X$ to Abelian groups. Now define:

$$\mathcal{F}(U) = \underset{{x\in U}}{\underleftarrow{\lim}} F(x)$$

If I take stalks and then do the above will I get back to the same sheaf? (Possibly after sheafication).

EDIT: Some details are missing. Whenever I'm taking limit of stalks, the category I'm taking the limit over is the poset of the points of the space. Where we have $x_0 \to x$ Iff $x$ is a generization of $x_0$ (i.e. if $x_0 \in \overline{\{x\}}$).

• @Hoot The field of fractions $K(X)=A_{\eta}$ where $\eta$ is the generic point - 0. Commented Dec 25, 2015 at 13:20
• I'm worried that your second display is going to fail if I cook up some non-Noetherian scheme where specialization/generalization doesn't tell the full topological story. Do you have a proof of this?
– Hoot
Commented Dec 25, 2015 at 14:47
• What are the transition maps between the $\mathscr F_x$ for varying $x$ that you use to define the limit?
– Remy
Commented Dec 25, 2015 at 19:02
• @Remy Added a more explicit construction Commented Dec 25, 2015 at 19:33
• @SaalHardali have you gotten any kind of answer for this question? How about posting it on MO? Commented Jan 13, 2016 at 14:33

First of all:

Let $$X=Spec A$$ be an affine integral scheme with structure sheaf $$\mathcal{O}_X$$.

We have: $$\mathcal{O}_{X,x}= \underset{{x\in X_f}}{ \underrightarrow{\lim}} \mathcal{O}_X(X_f)=\underset{{f \notin \mathfrak{p}_x}}{\underrightarrow{\lim}} A_f = \bigcup_{f \notin \mathfrak{p}_x} A_f$$ But we also have (I hope): $$\mathcal{O}_X(X_f)=A_f=\bigcap_{f\notin\mathfrak{p}_x\subset A} A_{\mathfrak{p}_x}=\underset{{f \notin \mathfrak{p}_x}}{\underleftarrow{\lim}} A_{\mathfrak{p}_x}=\underset{{x \in X_f}}{\underleftarrow{\lim}} \mathcal{O}_{X,x}$$

The first equality is false, because for any commutative ring $$A$$ with unit: $$$$\forall\mathfrak{p}\in\operatorname{Spec}A,\,\lim_{\overrightarrow{f\notin\mathfrak{p}}}A_f=A_{\mathfrak{p}}$$$$ and the second equality is partially true, that is: $$$$\mathcal{O}_X(D(f))=A_f=\bigcap_{\stackrel{\mathfrak{p}\in\operatorname{Spec}A}{f\notin\mathfrak{p}}}A_{\mathfrak{p}}$$$$ where I prefer the notation $$D(f)$$ for the open set $$$$\{x\in\operatorname{Spec}A\mid f(x)\neq0\}.$$$$ At most in general, the following lemma holds.

Lemma. Let $$(X,\mathcal{T})$$ be a topological space with topology $$\mathcal{T}$$ and $$\mathfrak{B}$$ a basis for $$\mathcal{T}$$; that is a system of open subsets of $$X$$ such that:

1. $$U,V\in\mathfrak{B}\Rightarrow U\cap V\in\mathfrak{B}$$,
2. every open subset of $$X$$ is a union of sets from $$\mathfrak{B}$$.

We can view $$\mathfrak{B}$$ as a category with objets its elements and the inclusions between the sets as morphisms.

Let $$\mathcal{O}:\mathfrak{B}\to\mathbf{C}$$ be a contravariant functor (or $$\mathfrak{B}$$-presheaf), where $$\mathbf{C}$$ is a category closed with respect to projective limits, such that $$\mathcal{O}$$ satisfies the sheaf conditions for coverings of type $$\displaystyle U=\bigcup_{i\in I}U_i$$, where $$\forall i\in I,\,U,U_i\in\mathfrak{B}$$.

Then $$\mathcal{O}$$ can be extended to a sheaf $$\overline{\mathcal{O}}$$ on $$X$$, where: $$$$\ \ \ \ \forall U\in\mathcal{T},\,\overline{\mathcal{O}}(U)=\lim_{\overleftarrow{V\in\mathfrak{B}\,\text{with}\,V\subseteq U}}\mathcal{O}(V)$$$$ and this extension is unique up to canonical isomorphism.

For a proof one can consult Bosch S. - Algebraic Geometry and Commutative Algebra, chapter 6, section 6, lemma 4.

Let $$(X,\mathcal{T})$$ be a topological space and let $$\mathcal{F}$$ be a sheaf on $$X$$ with values in a category $$\mathbf{C}$$ closed with respect to inductive and projective limits.

Let $$x,y\in X$$ such that $$x\in\overline{\{y\}}$$, or in other words: $$$$\forall U\in\mathcal{T},\,x\in U\Rightarrow y\in U;$$$$ then the following diagrams commute: $$$$\require{AMScd} \forall V\subseteq U\in\mathcal{T},x,y\in V,x\in\overline{\{y\}},\, \begin{CD} \mathcal{F}(U) @>r_{U,x}>> \mathcal{F}_x\\ @V{r^U_V}VV @VV{=}V\\ \mathcal{F}(V) @>>\dot\exists r_{V,x}> \mathcal{F}_x \end{CD}, \begin{CD} \mathcal{F}(U) @>r_{U,y}>> \mathcal{F}_y\\ @V{r^U_V}VV @VV{=}V\\ \mathcal{F}(V) @>>\dot\exists r_{V,y}> \mathcal{F}_y \end{CD},\dot\exists r_{y,x}:\mathcal{F}_x\to\mathcal{F}_y$$$$ and therefore $$(\mathcal{F}_x,r_{y,x})_{x,y\in X}$$ is a projective system in $$\mathbf{C}$$; where I get: $$\begin{gather} x\succcurlyeq y\iff x\in\overline{\{y\}}\,\text{or}\,x=y;\\ \forall U\in\mathcal{T},\,\mathcal{G}(U)=\lim_{\overleftarrow{x\in U}}\mathcal{F}_x. \end{gather}$$ Let $$\mathfrak{B}$$ be a basis of $$(X,\mathcal{T})$$, let $$V\subseteq U\in\mathfrak{B}$$, let $$x,y\in V$$ such that $$y\in\overline{\{x\}}\iff x\prec y$$; then one can consider the diagram $$$$\mathcal{G}(U)\stackrel{\displaystyle\left(r_y^U\right)^{\prime}}{\longrightarrow}\mathcal{F}_y\stackrel{\displaystyle r_{x,y}}{\longrightarrow}\mathcal{F}_x\stackrel{\displaystyle\left(r_x^V\right)^{\prime}}{\longleftarrow}\mathcal{G}(V)$$$$ by the universal property of $$\mathcal{G}(V)$$: $$$$\dot\exists r^U_V:\mathcal{G}(U)\to\mathcal{G}(V)\mid\left(r_x^V\right)^{\prime}\circ r^U_V=r_{x,y}\circ\left(r_y^U\right)^{\prime},$$$$ by definition $$\mathcal{G}$$ is a $$\mathfrak{B}$$-presheaf.

The $$\mathfrak{B}$$-sheaf axioms for $$\mathcal{G}$$ are equivalent to affirm that for any $$U\in\mathfrak{B}$$ and for any (open) covering $$\{U_i\in\mathfrak{B}\}_{i\in I}$$, $$\mathcal{G}(U)$$ is the equilizer of the diagram $$$$\prod_{i\in I}\mathcal{G}(U_i)\rightrightarrows\prod_{i,j\in I}\mathcal{G}(U_{ij})$$$$ where I get $$U_{ij}=U_i\cap U_j$$ and the double arrows is the categorical product (in $$\mathbf{C}$$) of the morphism $$r^{U_i}_{U_{ij}}$$ and $$r^{U_j}_{U_{ij}}$$.

By definition of $$\mathcal{G}$$: $$\mathcal{G}(U)$$ equalizes the previous diagram; let $$E$$ be the equalizer of the previous diagram, then: $$$$\forall i,j\in I,x_i\in U_i,y_{ij}\in U_{ij},\dot\exists\varphi_i:E\to\mathcal{F}_{x_i},\varphi_{ij}:E\to\mathcal{F}_{y_{ij}}$$$$ such that the $$\varphi_i$$'s and $$\varphi_{ij}$$'s commute opportunely with the $$r_{x_{ij},x_i}$$'s; by definition $$(E,\varphi_{ij})_{i,j\in I}$$ is a cone (in $$\mathbf{C}$$) on the projective system $$(\mathcal{F}_x,r_{y,x})_{x,y\in U}$$, by the universal properties of $$\mathcal{G}(U)$$ and $$E$$: they are canonically isomorphic, that is $$\mathcal{G}$$ is a $$\mathfrak{B}$$-sheaf; by previous lemma $$\mathcal{G}$$ is extendible to a sheaf on $$X$$.

Whithout confusion, I can continue to write $$\mathcal{G}$$ for both sheaves!

For any $$U\in\mathcal{T}$$, by previous reasoning, one can consider $$\mathcal{F}(U)$$ as a cone over the projective system $$(\mathcal{F}_x,r_{y,x})_{x,y\in U}$$; then by universal property of $$\mathcal{G}(U)$$, there exists a unique morphism $$\varphi_U:\mathcal{F}(U)\to\mathcal{G}(U)$$ such that it (opportunely) commutes whit the $$r_{y,x}$$'s; in particular, the data of $$\varphi_U$$'s defines a morphism $$\varphi:\mathcal{F}\to\mathcal{G}$$ of sheaves. In this way, one can define the canonical morphism $$\varphi_x:\mathcal{F}_x\to\mathcal{G}_x$$, for any $$x\in X$$.

For any $$x\in X$$, $$\left(\mathcal{F}_x,\left(r^U_x\right)^{\prime}\right)_{x\in U}$$ and $$\left(\mathcal{G}_x,r^U_x\right)_{x\in U}$$ are cocones for the inductive system $$\left(\mathcal{G}(U),r^U_V\right)_{x\in U,V}$$; by the couniversal property of $$\mathcal{G}_x$$, there exists a unique morphism $$\psi_x:\mathcal{G}_x\to\mathcal{F}_x$$; in particular, the data of $$\psi_x$$'s defines a morphism $$\psi:\mathcal{G}\to\mathcal{F}$$ of sheaves.

Using the couniversal property of the stalks of a sheaf, one can prove that: $$$$\forall x\in X,\,\psi_x\circ\varphi_x=Id_{\mathcal{F}_x},\varphi_x\circ\psi_x=Id_{\mathcal{G}_x};$$$$ in other words, $$\mathcal{F}$$ and $$\mathcal{G}$$ are canonical isomorphic sheaves on $$X$$ with values in $$\mathbf{C}$$. $$\Box\,(Q.E.D.)$$

• This is a very useful answer to this post. I can't seem to figure out why we have $\varphi_{x} \circ \psi_{x} = Id_{\mathcal{G}_{x}}$, though. Can you give me a hint or a good reference for this? It seems to me that it comes down to showing that $\varphi_{x} \circ r^{U}_{x} = \pi^{U}_{x}$, where $\pi^{U}_{x}$ is the canonical map from $\mathcal{G}(U)$ to the stalk $\mathcal{G}_{x}$ Commented Feb 7, 2022 at 17:57
• I use the Universal Property of a CoCone for an Inductive System: do you know this? Commented Feb 10, 2022 at 12:58
• Yeah, I know the property and used it precisely to define $\psi_{x}$, but wasn't able to show that $\varphi_{x} \circ \psi_{x} = Id_{\mathcal{G}_{x}}$. I proved that $\psi_{x} \circ \varphi_{x} = Id_{\mathcal{F}_{x}}$ by showing that $\psi_{x} \circ \varphi_{x} \circ \pi_{x}^{U} = \psi_{x} \circ \pi'^{U}_{x} \circ \varphi_{U} = r^{U}_{x} \circ \varphi_{x} = \pi_{x}^{U}$, which allows me to conclude the previous statement precisely because of the universality of the cocone associated with $\mathcal{F}$ and its stalk. Here, $\pi^{U}_{x}$ and $\pi'^{U}_{x}$ are the canonical maps in the cocones.. Commented Feb 10, 2022 at 15:15
• ... of $\mathcal{F}$ and $\mathcal{G}$ respectively Commented Feb 10, 2022 at 15:18
• But, for the other equality, I can only seem to get that $\varphi_{x} \circ \psi_{x} \circ \pi'^{U}_{x} \circ \varphi_{U} = \varphi_{x} \circ \psi_{x} \circ \varphi_{x} \circ \pi^{U}_{x} = \varphi_{x} \circ \pi^{U}_{x} = \pi'^{U}_{x} \circ \varphi_{U}$ (using the fact that $\psi_{x} \circ \varphi_{x} = Id_{\mathcal{F}_{x}}$), when really I would want to withdraw $\varphi_{U}$ from both sides Commented Feb 10, 2022 at 15:22