Last time I described two viewpoints on sheaves. In the first, a sheaf on a topological space is a special sort of presheaf

Namely, it’s one obeying the ‘sheaf condition’.

I explained this condition in Part 1, but here’s a slicker way to say it. Suppose is an open set covered by a collection of open sets Then we get this diagram:

The first arrow comes from restricting elements of to the smaller sets The other two arrows come from this: we can either restrict from to or restrict from to

The **sheaf condition** says that this diagram is an equalizer! This is just another way of saying that a family of are the restrictions of a unique iff their restrictions to the overlaps are equal.

In the second viewpoint, a sheaf is a bundle over

with the special property of being ‘etale’. Remember, this means that every point in has an open neighborhood that’s mapped homeomorphically onto an open neighborhood in

Last time I showed you how to change viewpoints. We got a functor that turns presheaves into bundles

and a functor that turns bundles into presheaves:

Moreover, I claimed actually turns presheaves into *etale spaces*, and actually turns bundles into *sheaves*. And I claimed that these functors restrict to an equivalence between the category of sheaves and the category of etale spaces:

What can we do with these ideas? Right away we can do two things:

• We can describe ‘sheafification’: the process of improving a presheaf to get a sheaf.

• We can see how to push forward and pull back sheaves along a continuous map between spaces.

I’ll do the first now and the second next time. I’m finding it pleasant to break up these notes into small bite-sized pieces, shorter than my actual lectures.

### Sheafification

To turn a presheaf into a sheaf, we just hit it with and then with In other words, we turn our presheaf into a bundle and then turn it back into a presheaf. It turns out the result is a sheaf!

Why? The reason is this:

**Theorem.** If we apply the functor

to any object, the result is a sheaf on

(The objects of are, of course, the bundles over )

Proving this theorem was a puzzle last time; let me outline the solution. Remember that if we take a bundle

and hit it with we get a presheaf called where is the set of sections of over and we restrict sections in the usual way, by restricting functions. But you can check that if we have an open set covered by a bunch of open subsets and a bunch of sections on the that agree on the overlaps these sections piece together to define a unique section on all of that restricts to each of the So, is a sheaf!

It follows that sends presheaves to sheaves. Since sheaves are a full subcategory of presheaves, any automatically sends any *morphism* of presheaves to a *morphism* of sheaves, and we get the **sheafification functor**

To fully understand this, it’s good to actually take a presheaf and sheafify it. So take a presheaf:

When we hit this with we get a bundle

Remember: any element of for any open neighborhood of gives a point over in all points over show up this way, and two such elements determine the same point iff they become equal when we restrict them to some sufficiently small open neighborhood of

When we hit this bundle with we get a sheaf

where is the set of sections of over This is the sheafification of

So, if you think about it, you’ll see this: to define a section of the sheafification of over an open set you can just take a bunch of sections of over open sets covering that agree when restricted to the overlaps.

**Puzzle.** Prove the above claim. Give a procedure for constructing a section of over given open sets covering and sections of over the that obey

### The adjunction between presheaves and bundles

Here’s one nice consequence of the last puzzle. We can always use the trivial cover of by itself! Thus, any section of over gives a section of over This is the key to the following puzzle:

**Puzzle.** Show that for any presheaf there is morphism of presheaves

Show that these morphisms are natural in so they define a natural transformation

Now, this is just the sort of thing we’d expect if were the left adjoint of Remember, when you have a left adjoint and a right adjoint you always have a ‘unit’

and a ‘counit’

where the double arrows stand for natural transformations.

And indeed, in Part 2 I claimed that is the left adjoint of But I didn’t prove it. What we’re doing now could be part of the proof: in fact Mac Lane and Moerdijk prove it this way in Theorem 2 of Section II.6.

Let’s see if we can construct the counit

For this I hand you a bundle

You form its sheaf of sections and then you form the etale space of that. Then you want to construct a morphism of bundles from your etale space to my original bundle.

Mac Lane and Moerdijk call the construction ‘inevitable’. Here’s how it works. We get points in over from sections of over open sets containing But you can just take one of these sections and evaluate it at and get a point in

**Puzzle.** Show that this procedure gives a well-defined continuous map

and that this is actually a morphism of bundles over Show that these morphisms define a natural transformation

Now that we have the unit and counit, if you’re feeling ambitious you can show they obey the two equations required to get a pair of adjoint functors, thus solving the following puzzle:

**Puzzle.** Show that

is left adjoint to

If you’re not feeling so ambitious, just look at Mac Lane and Moerdijk’s proof of Theorem 2 in Section II.6!

[…] I don’t think there’s anything especially sneaky about their argument. They do however use this: if you take a sheaf, and convert it into an etale space, and convert that back into a sheaf, you get back where you started up to natural isomorphism. This isomorphism is just the counit that I mentioned in Part 3. […]