I'm trying to understand this section of the Wikipedia article on sheaves. I think I get what a presheaf is, but there are parts of the locality and gluing axioms for sheaves that confuse me.
In my understanding, a presheaf just a contravariant functor from a category $\mathcal{C}$ to $\mathrm{Set}$, and that's a complete definition of a presheaf. However, the special case when $\mathcal{C}$ is the poset-category associated with the open sets of a topology is interesting, so we will consider just that special case for now.
On a very high level, checking whether the presheaf is a sheaf seems to amount to checking how it treats open covers of open sets and guaranteeing us something kind of like an existence and uniqueness property for functions defined locally on open covers… I think.
It is a little bit unclear how to check the locality rule, however. If I want two elements of $F(U)$ to be equal, is it sufficient that they be indistinguishable in one open covering? Or do they need to be indistinguishable in every open covering?
Also, in the gluing axiom, do we know that the gluing is unique by the axiom itself or is the uniqueness of the gluing a theorem rather than an axiom?
What follows is a quotation from the Wikipedia article and my attempt to understand the definition.
From Wikipedia, here is the locality axiom for sheaves
If $\mathcal{U}$ is an open covering of an open set $U$, and $s, t \in F(U)$
have the property $s|_{U_i} = t|_{U_i}$ for each set $U_i$ of the
covering, then $s=t$.
Here is the gluing axiom
If $\mathcal{U}$ is an open covering of an open set $U$, and if for
each $i \in I$ a section $s_i \in F(U_i)$ is given such that for each
pair $U_i, U_j$ of the covering sets the restrictions of $s_i$ and
$s_j$ agree on the overlaps $s_i|_{U_i \cap U_j} = s_j|_{U_i \cap U_j}$,
then there is a section $s \in F(U)$ such that $s|U_i = s_i$
for each $i \in I$.
First a word on notation, let $\tau^o$ be the open sets associated with a topological space $X$. This distinguishes it from $\tau^c$, the set of closed sets.
Let $f \mathop\square g$ be $g \circ f$.
Let a presheaf be the functor unique associated with the following four-place relation $R$.
$$ R \subset \tau^o \times \tau^o \times \mathrm{Set} \times \mathrm{Set} $$
$$ R(U, W, x, a) \; \text{holds} \;\; \textit{iff} \;\; \text{the restriction map from $U$ to $W$ sends $x$ to $a$} $$
Additionally, let the restriction map notation be defined in the following way.
$$\mathrm{res}_{V \leftarrow U}(s) = z \iff R(U, V, s, z)$$
Additionally, as a special case when $U$ is the domain, I define the following orthographic abbreviation
$$ s|_{V} \;\; \text{is an abbreviation for} \;\; \mathrm{res}_{V \leftarrow U}(s) $$
Let $F$ be a function defined the following way.
$$ V = F(U) \iff (\forall a \in V \mathop. R(U, U, a, a)) $$
Let $R'$ be a two-place function defined in the following way.
$$ R' : \tau^o \times \tau^o \to \mathrm{Set} $$
$$ R'(W, V) = \{ (a, b) : R(W, V, a, b) \} $$
$R$ satisfies the following axioms.
$$ R'(U, V) \mathop\square R'(V, W) = R'(U, W) \tag{101} $$
$$ R'(U, U) = \{ (a, a) : a \in F(U) \} \tag{105} $$
(101) is equivalent to saying that the functor associated with $R$ satisfies the composition law. (105) is saying that the restriction map from $U$ to $U$ is the identity map.
With that out of the way, here is my attempt to understand the sheaf rules.
Let $U$ be an element of $\tau^o$. Let $\bar{U}$ be an open covering of $U$ indexed by the index set $I$.
The locality (111,112) condition is defined as follows.
$$ \text{for all $i$ and $j$ in $I$} \; s|_{\bar{U}_i} = t|_{\bar{U}_j} \;\;\textit{implies}\;\; s = t \tag{111} $$
This is straightforwardly equivalent to
$$ (\forall i \in I \mathop. \forall a \mathop. \forall b \mathop. R(U, \bar{U}_i, s, a) \land R(U, \bar{U}_i, t, b) \to a = b) \to s =t \tag{112} $$
Let the gluing axiom be defined as follows.
Let $s$ be a function from a family of open sets $\bar{U}$, which is an open covering of $U$.
$$ s : \bar{U} \to \cup_{i \in I} F(\bar{U}_i) $$
If it is the case that
$$ s(\bar{U}_i) |_{\bar{U}_i \cap \bar{U}_j} = s(\bar{U}_j) |_{\bar{U}_i \cap \bar{U}_j} \;\; \text{for all $i$ and $j$ in $I$} \tag{121} $$
then it holds that
$$ \exists! e \in F(U) \mathop. e|_{\bar{U}_i} = s(\bar{U}_i) \;\; \text{for all $i$ in $I$} \tag{123} $$
As a side note, I may be misreading the Wikipedia article. I don't know whether the gluing axiom promises us a unique gluing or merely at least one gluing.
Getting back to the locality axiom (112) of a second, it's a little unclear to me how the scope of the variables is supposed to work.
$\bar{U}$ is an open covering of $U$, where $U$ is fixed outside the scope of the two well-formed formulas below.
(112) might be equivalent to (141) or (143) below.
Do we need to have only one open covering where $s$ and $t$ are equal under every restriction?
$$ \exists \bar{U} \mathop. (\forall i \in I \mathop. \forall a \mathop. \forall b \mathop. R(U, \bar{U}_i, s, a) \land R(U, \bar{U}_i, t, b) \to a = b) \to s =t \tag{141} $$
Or do we need every open covering to treat them the same way?
$$ \forall \bar{U} \mathop. (\forall i \in I \mathop. \forall a \mathop. \forall b \mathop. R(U, \bar{U}_i, s, a) \land R(U, \bar{U}_i, t, b) \to a = b) \to s =t \tag{143} $$
Best Answer
I don't understand what's confusing about what's written in Wikipedia.
The locality axiom says that if there is some open cover $\mathcal{U} = (U_i)_{i\in I}$ such that $s|_{U_i} = t|_{U_i}$ for all $i\in I$, then $s = t$. Of course, once you know $s = t$, then it follows that $s$ and $t$ agree upon restriction to every set in every open cover, since if $s = t$ and $V\subseteq U$ is open, then $s|_V = t|_V$.
The gluing axiom says nothing about uniqueness. But uniqueness follows from the locality axiom. If $s$ and $s'$ are two sections in $F(U)$ such that $s|_{U_i} = s_i$ and $s'|_{U_i} = s_i$ for all $i\in I$, then applying locality to the open cover $(U_i)_{i\in I}$, we conclude that $s = s'$.