# Sheafification of the constant presheaf

Let $$A$$ be an abelian group, and $$X$$ a topological space. Define the constant sheaf $$\mathcal{A}$$ on $$X$$ determined by $$A$$ as follows: for any open set $$U \subset X$$,

$$\mathcal{A}(U)=$$the group of continuous functions from $$U$$ to $$A$$, where $$A$$ is endowed with the discrete topology; with the usual restrictions we obtain a sheaf.

Moreover, the constant presheaf $$\mathcal{F}$$ of $$A$$ is defined by $$\mathcal{F}(U)=A$$ when $$U \neq \emptyset$$, again considering the usual restrictions we obtain a presheaf.

I am asked to show that the sheafification of $$\mathcal{F}$$ is $$\mathcal{A}$$.For this, I wanted to use the universal property of sheafification, and considered the application $$\theta :\mathcal{F} \to \mathcal{A}$$, $$\theta(U)(a):=s_a$$, where $$s_a$$ is the constant function on $$U$$, equal to $$a$$. I managed to show it is a morphism of sheaves. but given a sheaf $$\mathcal{G}$$ and a morphism $$\varphi: \mathcal{F} \to \mathcal{G}$$, I'm having trouble finding the unique morphism $$\psi :\mathcal{A} \to \mathcal{G}$$ s.t. $$\psi \theta =\varphi$$

• Are you familiar with the 'local homeomorphism' view of sheaves? Then $\psi$ becomes a continuous function between the total spaces, where $\mathcal A$ has all stalks isomorphic to $A$. Commented Dec 3, 2012 at 13:56

Let $\mathcal{F}'$ be the sheaf associated to $\mathcal{F}$ and let $\theta : \mathcal{F} \to \mathcal{F}'$ be the canonical morphism. Recall that $\theta$ induces an isomorphism between the stalks at every point (Liu, 2.2.14); hence the stalks of $\mathcal{F}'$ are $A$ everywhere. Now consider the morphism $\phi : \mathcal{F} \to \mathcal {A}$ which sends every element of $A$ to the correaponding constant map. The universal property of the sheaf associated to $\mathcal{F}$ gives a unique morphism $\psi : \mathcal{F}' \to \mathcal{A}$ through which $\mathcal{F}$ factors. Prove that this is an isomorphism by showing the induced maps on the stalks are isomorphisms (Liu, 2.2.12). For surjectivity, consider for any germ $f_x \in \mathcal{A}_x$ the germ of the section $f(x)$, and use the fact that continuous maps to a discrete space are locally constant.

• I don’t get the idea how to prove the “surjectivity” could you explain in more detail? Thanks
– user867836
Commented Sep 12, 2021 at 7:25

I'm late to the party, but here is the answer for reference's sake to the question of how to define the morphism $\psi$ in the verification of the universal property, as asked in the original question.

Let $f \in \mathcal{A}(U)$ for $U \subseteq X$ open. The key observation is that $\{f^{-1}(a)\}_{a \in A}$ forms a pairwise disjoint open cover of $U$. So it is enough to specify the value of $\psi(U)(f)|_{f^{-1}(a)}$ for each $a \in A$. The condition $\psi \circ \theta = \varphi$ forces us to define $\psi(U)(f)|_{f^{-1}(a)} = \varphi(U)(a)$. Define $\psi(U)(f)$ to be the section in $\mathcal{F}(U)$ obtained by gluing these together.

Here is an answer that goes into gory detail, which may be helpful for some people (it certainly was for me). If you do read my solutions, I suggest you go and rewrite them yourself actively.

So to recap, we have the presheaf $$\mathcal{F}$$ of constant functions and the sheaf $$\mathcal{A}$$ of locally constant functions for some abelian group $$A$$, i.e. $$\mathcal{A}(U)=\{f:U\to A \mid \forall x\in U \exists x\in V\subseteq U. f:V\to A \text{ constant}\}.$$ We want to show that $$\mathcal{F}^+\cong \mathcal{A}$$.

From the definition of $$\mathcal{F}^+$$ there exists a natural morphism $$\theta:\mathcal{F}\to \mathcal{F}^+$$ which induces an isomorphism of stalks $$\theta_x:\mathcal{F}_x\to \mathcal{F}_x^+$$. We also have a natural morphism $$\varphi:\mathcal{F}\to \mathcal{A}$$ that sends the element $$a\in A$$ to the constant function that maps to $$a$$. More explicitly, we have $$\mathcal{F}(U)=A$$ and so \begin{align*}\varphi_U:A &\to \mathcal{A}(U) \\ a &\mapsto (u\mapsto a) \end{align*} (If we are really pedantic we should explicitly state that this is well-defined, but it's easy to see that it is.)

From the universal property of sheafification we know get a unique morphism $$\psi:\mathcal{F}^+\to \mathcal{A}$$ such that $$\psi\circ \theta=\varphi$$. We now show that $$\theta$$ is an isomorphism and to do that, it suffices to show that $$\psi_x:\mathcal{F}_x^+ \to \mathcal{A}_x$$ is an isomorphism for all $$x$$.

Injective: The stalk $$\mathcal{F}_x^+ \cong \mathcal{F}_x\cong A$$ and so the map on stalks is $$\psi_x:A \to \mathcal{A}_x$$ given by $$\psi_x(a)=(u\mapsto a)_x$$. If $$\psi_x(a)=\psi_x(b)$$ then $$(u\mapsto a)_x = (u\mapsto b)_x$$ so that $$(u\mapsto a)=(u\mapsto b)$$ when restricted to some $$x\in W\subseteq U$$. Applying to $$x$$ gives $$a=b$$.

Surjective: As suggested by user314's answer, consider $$f_x\in \mathcal{A}_x$$. Because $$\psi_x\circ \theta_x=\varphi_x$$ it suffices to show that $$\varphi_x$$ is surjective now. As $$f:U\to A$$ is locally constant, there exists $$x\in W\subseteq U$$ such that $$f:W\to A$$ is constant. Now let $$a=f(x)$$. Then $$\varphi_x(a)=f_x$$.