# The Unapologetic Mathematician

## Sheaves

For the moment we will be more concerned with presheaves, but we may as well go ahead and define sheaves. These embody the way that not only can we restrict functions to localize them to smaller regions, but we can “glue together” local functions on small domains to define functions on larger domains. This time, let’s start with the fancy category-theoretic definition.

For any open cover $\{U_i\}_{i\in I}$ of an open set $U$, we can set up the following diagram:

$\displaystyle\mathcal{F}(U)\to\prod\limits_{i\in I}\mathcal{F}(U_i)\rightrightarrows\prod\limits_{(i,j)\in I\times I}\mathcal{F}(U_i\cap U_j)$

Let’s talk about this as if we’re dealing with a sheaf of sets, to make more sense of it. Usually our sheaves will be of sets with extra structure, anyway. The first arrow on the left just takes an element of $\mathcal{F}(U)$, restricts it to each of the $U_i$, and takes the product of all these restrictions. The upper arrow on the right takes an element of $\mathcal{F}(U_i)$ and restricts it to each intersection $U_i\cap U_j$. Doing this for each $U_i$ we get a map from the product over $i\in I$ to the product over all pairs $(i,j)$. The lower arrow is similar, but it takes an element in $\mathcal{F}(U_j)$ and restricts it to each intersection $U_i\cap U_j$. This may look the same, but the difference in whether the original set was the first or the second in the intersection makes a difference, as we shall see.

Now we say that a presheaf $\mathcal{F}$ is a sheaf if and only if this diagram is an equalizer for every open cover $U_i$. For it to be an equalizer, first the arrow on the left must be a monomorphism. In terms of sets, this means that if we take two elements $s\in\mathcal{F}(U)$ and $t\in\mathcal{F}(U)$ so that $s\vert_{U_i}=t\vert_{U_i}$ for all \$latex $U_i$, then $s=t$. That is, elements over $U$ are uniquely determined by their restrictions to any open cover.

The other side of the equalizer condition is that the image of the arrow on the left consists of exactly those products in the middle for which the two arrows on the right give the same answer. More explicitly, let’s say we have an $s_i\in\mathcal{F}(U_i)$ for each $U_i$, and let’s further assume that these elements agree on their restrictions. That is, we ask that $s_i\vert_{U_i\cap U_j}=s_j\vert_{U_i\cap U_j}$. If this is true for all pairs $(i,j)$, then the product $\left(s_i\right)_{i\in I}$ takes the same value under either arrow on the right. Thus it must be in the image of the arrow on the left — there must be some $s\in\mathcal{F}(U)$ so that $s\vert_{U_i}=s_i$. In other words, as long as the local elements $s_i\in\mathcal{F}(U_i)$ “agree” where their domains overlap, we can “glue them together” to give an element $s\in\mathcal{F}(U)$.

Again, the example to keep in mind is that of continuous real-valued functions. If we have a continuous function $f_U:U\to\mathbb{R}$ and another continuous function $f_V:V\to\mathbb{R}$, and if $f_U(x)=f_V(x)$ for all $x\in U\cap V$, then we can define $f:U\cup V\to\mathbb{R}$ by “gluing” these functions together over their common overlap: $f(x)=f_U(x)$ if $x\in U$, $f(x)=f_V(x)$ if $x\in V$, and it doesn’t matter which we choose when $x\in U\cap V$ because both functions give the same value there.

So, a sheaf is a presheaf where we can glue together elements over small domains so long as they agree when restricted to their intersections, and where this process defines a unique element over the larger, “glued-together” domain.

March 17, 2011 - Posted by | Topology

1. […] As ever, we want our objects of study to be objects in some category, and presheaves (and sheaves) are no exception. But, luckily, this much is […]

Pingback by Mappings Between Presheaves « The Unapologetic Mathematician | March 19, 2011 | Reply

2. For the left-most arrow in the first diagram to work, don’t the $U_i$ have to be subsets of $U$, which isn’t necessary for an open cover of the topological space, or am I once again missing something?

Comment by Avery Andrews | March 20, 2011 | Reply

3. oops open cover of $U$ in the topological space

Comment by Avery Andrews | March 20, 2011 | Reply

4. It’s not required for an open cover, but since $U$ is an open subspace we can always just pass to the intersection of the covering sets with $U$ itself. We still have an open cover of $U$, but all the sets are subsets of $U$.

Comment by John Armstrong | March 20, 2011 | Reply

5. I managed to think of this right after posting the question, but it’s nice to have it confirmed.

Comment by Avery Andrews | March 20, 2011 | Reply

6. […] Direct Image Functor So far our morphisms only let us compare presheaves and sheaves on a single topological space . In fact, we have a category of sheaves (of sets, by default) on . […]

Pingback by The Direct Image Functor « The Unapologetic Mathematician | March 21, 2011 | Reply

7. […] that we’ve talked a bunch about presheaves and sheaves in general, let’s talk about some particular sheaves of use in differential topology. Given a […]

Pingback by Sheaves of Functions on Manifolds « The Unapologetic Mathematician | March 23, 2011 | Reply

8. p 106-108 of http://folli.loria.fr/cds/1999/library/pdf/barrwells.pdf is a decent piece of side-reading for this, I think.

Comment by Avery D Andrews | March 27, 2011 | Reply

9. Nitpick: s/left/right/ in “upper arrow on the left”