Here’s a problem from the Sheaves section of Hartshorne. I’m having difficulty with the very last part…

*Background*. Consider the forgetful functor from the category of sheaves on to the category of presheaves on . Then if is a presheaf on , there is a universal morphism, say , from to this functor. is called the **sheaf associated to the presheaf ** or alternatively its **sheafification**.

Let be an abelian group. Equip it with the discrete topology. The corresponding **constant sheaf**, denoted , is the one which maps each open set to the group of continuous functions , where is equipped with the discrete topology. The restriction maps are the obvious ones.

*Problem*. Let be an abelian group, and define the **constant presheaf **associated to on the topological space to be the presheaf for all , with restriction maps the identity. Show that the constant sheaf defined in the text is the sheaf associated to this presheaf.

*Solution*. Call this presheaf . We need to find a morphism with the property that for any morphism , there is a unique morphism such that .

Hence, let be an open set. We need a morphism of abelian groups . Recall that was the group of all continuous maps of into . The natural choice seems to be to send to the constant function . Let us verify that really is a morphism of presheaves: suppose we have an inclusion . Then the statement that the appropriate diagram commutes is the same as saying “the restriction to , of the constant function on , is the same as the constant function on ”, which is clear. So is a morphism of presheaves.

Before proceeding, we note that if is discrete then any continuous map is locally constant, in the sense that each admits some neighbourhood on which is constant. Locally constant maps are constant on each connected component.

Finally, suppose is a morphism of presheaves, i.e. for each open set we have a morphism of abelian groups . Define, for each open , a morphism as follows. If is connected, then since any continuous is constant (say for ), we may define . If is not connected, then we should have where are the connected components of , and then it should be clear where to go from here.

However, it seems like to finish, we instead want — we want to decompose any continuous as a finite sum of constant functions, and then the way to define will be clear. My question is: how do we know this is possible? What about, say, the function given by ?

**EDIT**: Maybe one should look at the maps induced on the stalks by , since to figure out what should be, it seems sufficient to only have local data about , which would certainly make our life easy since is locally constant…

I don’t understand this sentence, in your definition of the sheafification: Then if is a presheaf on , there is a universal morphism, say , from to this functor. I think this is basically supposed to be saying that the sheafification is (a functor from presheaves to sheaves and) left adjoint to the forgetful functor from sheaves to presheaves, right? Also, is it obvious that the sheafification always exists? I think there are general ways to ensure that adjoint functors exist (see: adjoint functor theorem lol) but I can’t really make sense of what these conditions say in general.

For the problem: You’re on the right track with your last comment (the edit). To define , first cover with open sets where is constant on each . Then you can use applied to the constant value that takes on to get an element of . Then use the sheafy-ness of to glue these together into an element of . Then it’s just a straightforward matter of checking that the resulting map is actually well-defined (doesn’t depend on choices of covering) and gives a morphism of sheaves, and that factors through the way you want it to.

The key idea is that you can cover an open set with smaller ones on which you have good behaviour, working things out on those, and then glue everything together at the end. The ability to do this is exactly why sheaves are the “correct” definition.

To address your first paragraph: if you unravel the definition of “universal morphism from an object to a functor”, I’m just noting (without proof) that there is an initial object in the “comma category” , where is the forgetful functor from sheaves to presheaves. Proposition 1 of III.2 in Mac Lane suggests there might be some sense in which what you said about adjoint functors is equivalent, but I don’t know enough about the latter yet to tell…

Okay that makes sense. The statement that the sheafification is left adjoint to to is a little bit stronger. This basically unpacks to saying that exists for all presheaves , and that is a functor (in addition to the definition of already given).

Also I totally forget how to make LaTeX display properly here. I could’ve sworn it was either $latex…$ or $\latex…$ but neither of those seem to work now.

It’s $latex…$, except you need a space between the word “latex” and the code.