Sheaves

In Presheaves, we discussed functions “living on” the open subsets of a topological space. In particular, we discussed an example of “opposite” or “contravariant” behavior of the sets of functions on the open subsets of a topological space. For an inclusion $V\subseteq U$ of open subsets of a topological space $X$ we had a function $\rho_{UV}: \mathcal{F}(U)\rightarrow \mathcal{F}(V)$ between the sets of functions on these open subsets.

The topological space itself can be “built up” in some way from its open subsets (we will later clarify in what sense) and we now want to look at how we want the sets of functions on these open subsets to “patch together” to form the set of functions on the topological space. In other words, we want to use the “local” information that we have to obtain something “global”.

Now we want to show what we mean by “building up” the topological space from its open subsets. We quote here from the book Topology by James R. Munkres:

A collection $\mathcal{A}$ of subsets of a space is said to cover $X$, or to be a covering of $X$, if the union of the elements of $\mathcal{A}$ is equal to $X$. It is called an open covering of $X$ if its elements are open subsets of $X$.

Now we quote from the book Algebraic Topology by Robin Hartshorne the conditions that make a presheaf a sheaf:

A presheaf $\mathcal{F}$ on a topological space $X$ is a sheaf if it satisifes the following supplementary conditions:

(3) if U is an open set, if $\{V_{i}\}$ is an open covering of $U$, and if $s\in\mathcal{F}(U)$ is an element such that $s|_{V_{i}}=0$ for all $i$, then $s=0$.

(4) if $U$ is an open set, if $\{V_{i}\}$ is an open covering of $U$, and if we have elements $s\in\mathcal{F}(V_{i})$ for each $i$, with the property that for each $i,j$$s_{i}|_{V_{i}\cap V_{j}}=s_{j}|_{V_{i}\cap V_{j}}$, then there is an element $s\in\mathcal{F}(U)$ such that $s|_{V_{i}}=s_{i}$ for each $i$.

Note condition (3) implies that $s$ is unique.

We recall our example in Presheaves. There we had as our topological space the complex plane $\mathbb{C}$ equipped with the Zariski topology. We saw that the sets of functions of the form $\frac{f}{g}$, where $f$ and $g$ are polynomials (again all functions discussed in this post will be of this form), on the open subsets of $\mathbb{C}$ form a presheaf. We will continue using this example to show what it means for a presheaf to be a sheaf.

Condition (3) above (the weird numbering convention is because the first two conditions are for presheaves, as was quoted in Presheaves) means that if we have a function which, when “restricted” to an open subset is $0$ for all the open subsets whose union is $\mathbb{C}$, then this function is the function $0$ on $\mathbb{C}$. For instance our open subsets might be $\mathbb{C}-\{0\}$ and $\mathbb{C}-\{1\}$; their union is $\mathbb{C}$. Together they form an open covering of $\mathbb{C}$. For condition (3) to be satisfied, if a function on $\mathbb{C}$ restricts to $0$ on both $\mathbb{C}-\{0\}$ and $\mathbb{C}-\{1\}$, then it must be the function $0$ on $\mathbb{C}$.

Condition (4) above means that if we have two functions on two open subsets that restrict to the same function on the intersection of these two open subsets, and if this is true for all pairs of open subsets, then these functions are restrictions of a function on $\mathbb{C}$. We consider again $\mathbb{C}-\{0\}$ and $\mathbb{C}-\{1\}$. We know that their union is $\mathbb{C}$; meanwhile their intersection is $\mathbb{C}-\{0,1\}$. For condition (4) to be satisfied, if we have functions on $\mathbb{C}-\{0\}$ and $\mathbb{C}-\{1\}$ which restrict to the same function on $\mathbb{C}-\{0,1\}$, then the functions on $\mathbb{C}-\{0\}$ and $\mathbb{C}-\{1\}$ must be the restrictions of a function on $\mathbb{C}$.

These functions that we have discussed are called regular functions on the open subsets of $\mathbb{C}$, and they do form a sheaf called the sheaf of regular functions on $\mathbb{C}$.

To see what makes a sheaf so special, it is perhaps more informative to look at an example of a presheaf which is not a sheaf. Let our topological space $X$ be the set with two elements $p$ and $q$, equipped with the discrete topology; in other words, every subset is declared to be an open set. There are only four of them, the entire set $X=\{p,q\}$, $\{p\}$, $\{q\}$, and the empty set $\varnothing$.

To construct a presheaf on $X$, we need to specify a set $\mathcal{F}(U)$ for every open subset $U$ of $X$ and restriction maps $\rho_{UV}:U\rightarrow V$ for every inclusion of open subsets $V\subseteq U$of $X$.

We choose $\mathcal{F}(U)=\mathbb{Z}$ for all open sets $U$ of $X$ except for the empty set, where we choose $\mathcal{F}(\varnothing)=0$. We choose identity functions from $\mathbb{Z}$ to $\mathbb{Z}$ for our restriction maps except for the restriction map to the empty set, where we choose instead the constant function that sends every element to $0$. We now check to see if this presheaf forms a sheaf.

First, we must show that if we have a section $s\in \mathcal{F}(X)$ which gets sent under the restriction maps to $0\in \mathcal{F}(\{p\})$ and $0\in \mathcal{F}(\{q\})$, then $s=0\in \mathcal{F}(X)$. Our restriction maps are just identity functions, so it is only the section $0\in \mathcal{F}(X)$ which restricts to $0$ on $\mathcal{F}(\{p\})$ and $\mathcal{F}(\{q\})$.

Second, we must show that if we have two sections over two open subsets of $X$ which get sent by the restriction map to the same section over the intersection of the two open subsets, and this holds for all open subsets that make up an open covering of $X$, then these two sections must be restrictions of a section over $X$. Our presheaf does not satisfy this condition.

To see why, we consider $\mathcal{F}(\{p\})$ and $\mathcal{F}(\{q\})$. The intersection of $\{p\}$ and $\{q\}$ is the empty set $\varnothing$. The restriction map to $\mathcal{F}(\varnothing)$ is always the constant function to the section $0\in \mathcal{F}(\varnothing)$, so any two sections, say $m\in \mathcal{F}(\{p\})$ and $n\in \mathcal{F}(\{q\})$, get sent to the same section $0\in \mathcal{F}(\varnothing)$.

Under the condition, these two sections must be restrictions of a section $s\in \mathcal{F}(X)$. But the restriction map is the identity function; this means that$s=m$ and $s=n$ as integers. But we specified earlier that $m$ and $n$ could be any two sections of $\mathcal{F}(\{p\})$ and $\mathcal{F}(\{q\})$, so they could be different, $m\neq n$, and it would be impossible for them to be both be equal to $s$. In other words, they cannot both be restrictions of the same section $s\in \mathcal{F}(X)$.

So now we see that our presheaf is not a sheaf. If, however, instead of setting $\mathcal{F}(X)=\mathbb{Z}$, we set $\mathcal{F}(U)=\mathbb{Z}\oplus \mathbb{Z}$, where $\mathbb{Z}\oplus\mathbb{Z}$ is a group, whose underlying set is the set of pairs of integers, with the law of composition just addition on each “component”, i.e. $(a, b)+(c, d)=(a+b, c+d)$, and the restriction maps to $\mathcal{F}(\{p\})$ and $\mathcal{F}(\{q\})$ (both still equal to $\mathbb{Z}$) the “projection maps” $\pi_{1}$ and $\pi_{2}$ from $\mathbb{Z}\oplus\mathbb{Z}$ to $\mathbb{Z}$ which send the pair $(m,n)\in \mathbb{Z}\oplus\mathbb{Z}$ to $m\in\mathbb{Z}$ and $n\in\mathbb{Z}$ respectively, we can, in fact, obtain a sheaf, an example of what is called a constant sheaf.

The concept of a sheaf is essential to modern algebraic geometry, which studies, among other things, varieties, or shapes that can be described by polynomial equations, together with the functions that “live on” them, which likewise can usually be described in terms of polynomials. More on sheaves, including the particular examples we have introduced here, can be found in the references below.

References:

Sheaf on Wikipedia

Constant Sheaf on Wikipedia

Algebraic Geometry by Andreas Gathmann

The Rising Sea: Foundations of Algebraic Geometry by Ravi Vakil

Algebraic Topology by Robin Hartshorne

Sheaves in Geometry and Logic: A First Introduction to Topos Theory by Saunders Mac Lane and Ieke Moerdijk

Topology by James R. Munkres