## Constant presheaf

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 $X$ to the category of presheaves on $X$. Then if $\mathscr{F}$ is a presheaf on $X$, there is a universal morphism, say $\theta : \mathscr{F} \to \mathscr{F}^+$, from $\mathscr{F}$ to this functor. $\mathscr{F}^+$ is called the sheaf associated to the presheaf $\mathscr{F}$ or alternatively its sheafification.

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

Problem. Let $A$ be an abelian group, and define the constant presheaf associated to $A$ on the topological space $X$ to be the presheaf $U \mapsto A$ for all $U \neq \varnothing$, with restriction maps the identity. Show that the constant sheaf $\mathscr{A}$ defined in the text is the sheaf associated to this presheaf.

Solution. Call this presheaf $\mathscr{F}$. We need to find a morphism $\theta : \mathscr{F} \to \mathscr{A}$ with the property that for any morphism $\varphi : \mathscr{F} \to \mathscr{G}$, there is a unique morphism $\psi : \mathscr{A} \to \mathscr{G}$ such that $\varphi = \psi \circ \theta$.

Hence, let $U \neq \varnothing$ be an open set. We need a morphism of abelian groups $\theta(U) : \mathscr{F}(U) = A \to \mathscr{A}(U)$. Recall that $\mathscr{A}(U)$ was the group of all continuous maps of $U$ into $A$. The natural choice seems to be to send $a \in A$ to the constant function $a \in \mathscr{A}(U)$. Let us verify that $\theta$ really is a morphism of presheaves: suppose we have an inclusion $V \subseteq U$. Then the statement that the appropriate diagram commutes is the same as saying “the restriction to $V$, of the constant function $a$ on $U$, is the same as the constant function $a$ on $V$”, which is clear. So $\theta$ is a morphism of presheaves.

Before proceeding, we note that if $Y$ is discrete then any continuous map $f : X \to Y$ is locally constant, in the sense that each $x \in X$ admits some neighbourhood on which $f$ is constant. Locally constant maps are constant on each connected component.

Finally, suppose $\varphi : \mathscr{F} \to \mathscr{G}$ is a morphism of presheaves, i.e. for each open set $U$ we have a morphism of abelian groups $\varphi(U) : \mathscr{F}(U) = A \to \mathscr{G}(U)$. Define, for each open $U$, a morphism $\psi(U) : \mathscr{A}(U) \to \mathscr{G}(U)$ as follows. If $U$ is connected, then since any continuous $f : U \to A$ is constant (say $f \equiv a$ for $a \in A$), we may define $\psi(U)(f) = \varphi(U)(a)$. If $U$ is not connected, then we should have $\mathscr{A}(U) \cong \prod \mathscr{A}(U_i)$ where $\{ U_i \}$ are the connected components of $U$, and then it should be clear where to go from here.

However, it seems like to finish, we instead want $\mathscr{A}(U) \cong \bigoplus \mathscr{A}(U_i)$ — we want to decompose any continuous $f : U \to A$ as a finite sum of constant functions, and then the way to define $\psi(U)(f)$ will be clear. My question is: how do we know this is possible? What about, say, the function $\mathbb{R} \setminus \mathbb{Z} \to \mathbb{Z}$ given by $x \mapsto \lfloor x \rfloor$?

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

just another guy trying to make the diagrams commute.
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### 5 Responses to Constant presheaf

1. Erik Crevier says:

I don’t understand this sentence, in your definition of the sheafification: Then if $\mathcal{F}$ is a presheaf on $X$, there is a universal morphism, say $\theta:\mathcal{F}\to\mathcal{F}^+$, from $\mathcal{F}$ 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 $\psi(U)(f)$, first cover $U$ with open sets $V_i$ where $f$ is constant on each $V_i$. Then you can use $\phi(V_i)$ applied to the constant value that $f$ takes on $V_i$ to get an element of $\mathcal{G}(V_i)$. Then use the sheafy-ness of $\mathcal{G}$ to glue these together into an element of $\mathcal{G}(U)$. 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 $\phi$ factors through $\psi$ 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.

• mlbaker says:

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” $\mathcal{F} \downarrow U$, where $U$ 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…

2. Erik Crevier says:

Okay that makes sense. The statement that the sheafification is left adjoint to to $U$ is a little bit stronger. This basically unpacks to saying that $\mathcal{F}^+$ exists for all presheaves $\mathcal{F}$, and that $\mathcal{F}\mapsto\mathcal{F}^+$ is a functor (in addition to the definition of $\mathcal{F}^+$ already given).

3. Erik Crevier says:

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.

• mlbaker says:

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