# Equality of restrictions in the definition of a sheaf by Rosiak

(Definition of a sheaf) Assume given $$X$$ a topological space, with $$\mathcal{O}(X)$$ its partial order category of open sets, and $$F \colon \mathcal{O}(X)^{\mathrm{op}} \to \mathrm{Set}$$ a presheaf. Then, given an open set $$U \subset X$$ and a collection $$\{U_i \}_{i \in I}$$ of open sets covering $$U = \bigcup_{i \in I} U_i$$, we can define the following sheaf condition.

Given a family of sections $$s_1, \dotsc, s_n$$ where each $$s_i \in F(U_i)$$ is a value assignment (section) over $$U_i$$, whenever we have that for all $$i, j$$, $$s_i|_{U_i \cap U_j } = s_j|_{U_i \cap U_j} \,,$$ then there exists a unique value assignment (section) $$s \in F(U)$$ such that $$s|_{U_i} = s_i$$ for all $$i$$.

Note: Given $$s ∈ F(U)$$, it is common to denote its restriction to $$V$$ by $$s|_V$$, that is, $$ρ_V^U \colon F(U) \to F(V)$$ takes $$s \mapsto s|_V$$ for each $$s ∈ F(U)$$,

Sheaf Theory through Examples by Rosiak, Chapter 5, Definition 122

In the above definition, what does the following condition mean? $$s_i|_{U_i \cap U_j } = s_j|_{U_i \cap U_j}$$ If $$s_i$$ are element in the set which functor takes open set to, what does it mean for two elements to be equal in the restriction?

For example, if I say $$3 = 1 + 2$$, that is the same whether I take the set which the numbers come from as natural, rational, real, etc.

As the note mentions, given $$V \subseteq U$$, there is a map $$\rho_V^U : F(U) \to F(V)$$ and for $$s \in F(U)$$, the image $$\rho_V^U(s)$$ is often denoted $$s|_V$$.

As $$U_i\cap U_j \subseteq U_i$$ and $$U_i\cap U_j \subseteq U_j$$, there are restriction maps $$\rho_{U_i\cap U_j}^{U_i} : F(U_i) \to F(U_i\cap U_j)$$ and $$\rho_{U_i\cap U_j}^{U_j} : F(U_j) \to F(U_i\cap U_j)$$. So for $$s_i \in F(U_i)$$ and $$s_j \in F(U_j)$$, the condition $$s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}$$ means

$$\rho^{U_i}_{U_i\cap U_j}(s_i) = \rho^{U_j}_{U_i\cap U_j}(s_j).$$

• Given the question is this really a helpful answer? Commented Jan 3 at 0:27
• @Alper: I think so. The OP asked what the condition $s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}$ means, so I unwrapped the notation to give a precise statement in terms of the structure of the presheaf $F$. I guess you read it as a question of intuition instead, which is another reasonable interpretation. Commented Jan 4 at 0:47

I take the condition to mean that on the same area (the restriction on the intersection) two different value assignments are not allowed to contradict each other, i.e. they have to agree.