# Gluing sheaves - can we realize $\mathcal{F}(W)$ as some kind of limit?

Let $X$ be a topological space and $\{U_i\}$ and open cover for $X$. Suppose we have sheaves $\mathcal{F}_i$ on $U_i$ and for each $i,j$ an isomorphism $\varphi_{ij} : \mathcal{F}_i|_{U_i \cap U_j} \to \mathcal{F}_j|_{U_i \cap U_j}$ such that $\varphi_{ii} = id$ and $\varphi_{ik} = \varphi_{jk} \circ \varphi_{ij}$ on $U_i \cap U_j \cap U_k$. I want to define some kind of sheaf $\mathcal{F}$ on $X$; to do this we consider

$$\mathcal{F}(W) := \Big\{(s_i)_{i \in I}, s_i \in \mathcal{F}_i(W \cap U_i) : \varphi_{ij}(s_i|_{W \cap U_i \cap U_j}) = s_j|_{W \cap U_i \cap U_j} \Big\}.$$

My question is: Can we realise $\mathcal{F}(W)$ as some kind of limit of a diagram, or yet as an equalizer of two maps? I ask this because I want to show that this $\mathcal{F}$ is a sheaf. At the moment it seems to me that $\mathcal{F}(W)$ is very close to being some kind of inverse limit", but I don't know exactly what it is.

• You are exactly right, the restriction maps give you an inverse system, and so the "glued" sheaf is just the inverse limit of this system. Commented Jul 30, 2013 at 15:39
• @RobertAuffarth Not exactly. The problem is that if we want the abelian groups in question to be $\mathcal{F}_i(W\cap U_i) =: A_i$ then what is the morphism from $A_j$ to $A_i$ say? It cannot be $\varphi_{ij}$...
– user38268
Commented Jul 30, 2013 at 15:47
• You're right, sorry about that! Commented Jul 31, 2013 at 1:04
• $\phi_{ii}=1$ is superfluous, it follows from the rest. Commented Feb 19, 2014 at 2:58

To answer your question, the sheaf $\mathscr{F}$ you are trying to construct can be described in the category $\mathbf{Sh}(X)$ as the limit of a certain diagram. We will consider sheaves on an open subset $U \subseteq X$ as sheaves on $X$ by the usual direct image construction.

• The diagram has one vertex for every ordered pair of elements in the open cover; the object at the vertex $(i, j)$ is the sheaf $\mathscr{F}_i |_{U_i \cap U_j}$.
• We have an edge $(i, i) \to (i, j)$ for every ordered pair $(i, j)$; the corresponding morphism $\mathscr{F}_i \to \mathscr{F}_i |_{U_i \cap U_j}$ is the restriction map.
• We have an edge $(i, j) \to (j, i)$ for every ordered pair $(j, i)$; the corresponding morphism $\mathscr{F}_i |_{U_i \cap U_j} \to \mathscr{F}_j |_{U_j \cap U_i}$ is the isomorphism $\varphi_{i,j}$.

The cocycle condition guarantees that all the triangles appearing in the diagram commute. Thus we have a well-defined diagram in $\mathbf{Sh}(X)$ and we can take its limit.

Amusingly, there is another sense in which gluing sheaves is a limit construction: the category $\mathbf{Sh}(X)$ itself is a bicategorical limit of a (pseudo)commutative diagram involving all the categories $\mathbf{Sh}(U_i)$, $\mathbf{Sh}(U_i \cap U_j)$, and $\mathbf{Sh}(U_i \cap U_j \cap U_k)$. In fancy language, we say that $\mathbf{Sh}(-)$ is a stack for the canonical topology on the site of open subsets of $X$.

• Dear Zhen, thanks for your answer. 1) Is it true that arbitrary limits exist in $\textbf{Sh}(X)$? Also, is it true that the sheaf $\mathcal{F}$ I produced above is the limit of your diagram in your answer?
– user38268
Commented Aug 1, 2013 at 10:06
• Yes on both counts. Commented Aug 1, 2013 at 10:18
• Zhen Lin, Can this bicategorical limit be expressed as a strict 1-limit of categories? For the stack of sheaves on a topological space, do we really need higher limits, or is 1-category theory enough? Commented Nov 16, 2014 at 21:58
• Yes, but that involves many technical details. (In the first place, what we have is not a commutative diagram but rather a pseudocommutative diagram.) Commented Nov 16, 2014 at 22:10
• Good to know, because I am in the process of checking it for myself. Thanks. Commented Nov 16, 2014 at 23:12