Sheafification of constant presheaf

Question

I'm reading about sheafification but it's too abstract for me. I'm trying to work out the example of sheafification of a constant presheaf and my advisor gives me this example: Let $X$ be a topological space and $A$ be an abelian group. Let $\mathfrak{A}^+$ be the presheaf of continuous functions, i.e. $$\mathfrak{A}^+(U)=\{\alpha : U → A \text{ continuous}\}$$ where $A$ has the discrete topology, for each open set $U$. There are 3 things he wants me to show:

(1) $\mathfrak{A}^+$ is a sheaf of abelian groups.

(2) All stalks $\mathfrak{A}^+_x=A$.

(3) The natural map $f : \mathfrak{A} → \mathfrak{A}^+$ from the constant sheaf induces an isomorphism on stalks

(1) is easy since a function is $0$ iff it is $0$ locally. I think (2) and (3) should be easy since we do not have much information, but strangely I don't know how to use them.

In (2), I know the maps $\mathfrak{A}^+(U)\to \mathfrak{A}^+(V)$ when $V\subseteq U$ is just restriction. By definition, the stalk $\mathfrak{A}^+_x=\varinjlim_{x\in U} \mathfrak{A}^+(U)$. But how do you convert the definition (group of continuous functions) to $A$? I even feel like we lack information here since we do not even know the structure of open sets in $X$, and I do not see any natural map from the direct limit to $A$.

(3) is even more ambiguous for me. Say $U$ is open. How do you naturally map an element in $A=\mathfrak{A}$, to a function $U\to A$ in $\mathfrak{A}^+$? I know that the maps satisfy a diagram, so $f_U(a)|_V=f_V(a)$ where $V\subseteq U$ are open sets. According to (2), the stalks of both are $A$, but I don't think the induced map would be the identity, so it makes it harder for me to understand why this is an iso?

I'm just a beginner in Algebraic Geometry so these questions may be obvious. Thank you.

Answer 1

(1) Much more generally (and therefore much easier!):
Given an arbitrary topological space $Y$ there is a sheaf $\mathcal C_Y$ on $X$ characterized by the requirement that for any open subset $U\subset X$ we have $\mathcal C_Y(U)=\mathcal C(U,Y)$, the set of continuous functions $U\to Y$.
Your $ \mathfrak A^+$ is exactly the sheaf $\mathcal C_A$, where $A$ is endowed with the discrete topology.

(2) We have a morphism of abelian groups $i:\mathcal C_{A,x}\to A$ defined as follows: given a germ $\gamma\in \mathcal C_{A,x}$ we represent it by some continuous function $g:U\to A$ and we define $i(\gamma)=g(x)$.
This morphism is an isomorphism since it has as explicit inverse the group morphism $j:A\to \mathcal C_{A,x}$ sending $a\in A$ to $j(a)= \operatorname [{const}(a)]_x$, the germ at $x$ of the constant function $\operatorname {const}(a): X\to A$ sending any point of $X$ to $a$.

(3) For any neighbourhood $U\subset X$ of $x$ the map $\mathfrak A(U)=A \to \mathcal C_A(U): a\mapsto \operatorname {const}(a)$ is clearly injective, and thus taking direct limits the map $f: \mathfrak A_x\to \mathcal C_{A,x}$ is also injective.
Surjectivity is easy too: in the notation of (b) a germ $\gamma=[g]_x\in \mathcal C_{A,x}$ is the image under $f$ of the germ $[\operatorname {const}(g(x))]_x\in \mathfrak A_x$.
And so we have proved that the morphism $f: \mathfrak A_x\to \mathcal C_{A,x}$ is indeed bijective.