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. $\psi \theta =\varphi$

Solution 1:

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.

Solution 2:

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.