$\mathcal{F}(X)\to\mathcal{F}(U)$ bijective for all $U$ implies $\mathcal{F}$ is locally constant I have been trying to solve the Exercise 2.13 from "Algebraic Geometry I" by Görtz and Wedhorn (page 63). It is not homework or anything, just for my own practice. For now, I want to ask about part (a). The problem is the following:
Problem. Let $X$ be irreducible. Show that the following properties for a presheaf $\mathcal{F}$ on $X$ are equivalent: 
(i) For every non-empty open subset $U\subset X$, the restriction map $\mathcal{F}(X)\to\mathcal{F}(U)$ is bijective. 
(ii) $\mathcal{F}$ is a constant sheaf on $X$. 
(iii) $\mathcal{F}$ is a locally constant sheaf on $X$.
First of all, I would like to ask a clarifying question:

When the authors says "$\mathcal{F}$ is a constant sheaf", they really mean it is a constant sheaf, and not just isomorphic (as sheaves) to a constant sheaf, right?

Okay, so since $X$ is irreducible, every non-empty open set is connected. In particular, every constant presheaf on $X$ is constant sheaf. How would I show that (i)$\Rightarrow$ (ii) or even (i)$\Rightarrow$ (iii)? I know that the condition of (i) automatically implies that $\mathcal{F}$ is a sheaf as well. Also, if $\mathcal{F}(X)\to \mathcal{F}(U)$ is bijective, it is pretty straightforward to see that $\mathcal{F}(U)\to\mathcal{F}(V)$ is bijective for every $\emptyset\neq V\subseteq U$. As a result, the condition (i) implies that the stalk $\mathcal{F}_{x}$ is bijective to $\mathcal{F}(X)$. The problem is that these are merely bijections! How do I show that if I pick $U$ "small enough", the restriction maps $\mathcal{F}(U)\to\mathcal{F}(V)$ will be all identity maps? This is, after all, the definition of the constant sheaf. 
I would appreciate any help on the problem! 
 A: Thanks to stewbasic's comment above, I realize that part (ii) actually means "$\mathcal{F}$ is isomorphic to the constant sheaf". Similarly, part (iii). Anyways, so let me try to sketch the direction (i) $\Rightarrow$ (ii). In the solution below, I will use the notation $\operatorname{res}_{\mathcal{F}}^{U, V}$ to denote the restrictions map for the sheaf $\mathcal{F}$. While it is somewhat cumbersome, it helps me keep track of which sheaf we are working on (as there will be two sheaves below).
Assume the condition (i). Then for every non-empty open set $U\subset X$, we have a bijection $g_{U}: \mathcal{F}(X)\to\mathcal{F}(U)$. If $V\subseteq U\subseteq X$, we have 
$$
g_{V}: \mathcal{F}(X)\overset{g_{U}}{\longrightarrow}\mathcal{F}(U)\overset{\operatorname{res}_{\mathcal{F}}^{U, V}}{\longrightarrow} \mathcal{F}(V)
$$
that is, $g_{V} = \operatorname{res}^{U, V}_{\mathcal{F}}\circ g_{U}$. Consequently, the restriction map $\operatorname{res}^{U, V}_{\mathcal{F}}: \mathcal{F}(U)\to\mathcal{F}(V)$ is given by $\operatorname{res}^{U, V}_{\mathcal{F}} = g_{V}\circ g_{U}^{-1}$. Next, let $\mathcal{G}$ be the constant sheaf on $X$ with value $\mathcal{F}(X)$. Since $X$ is irreducible, every open set of $X$ is connected, so the definition of $\mathcal{G}$ amounts to declaring $\mathcal{G}(U)=\mathcal{F}(X)$ for every non-empty open set $U\subseteq X$, and the restriction maps are identity maps. Define a morphism of sheaves as follows. For each open set $\emptyset\neq U\subseteq X$, we define
$$
\varphi_{U}: \mathcal{F}(U) \to \mathcal{G}(U)
$$
by $\varphi_{U}(s) = g^{-1}_{U}(s)\in \mathcal{F}(X)=\mathcal{G}(U)$. We need to check that it is indeed a morphism of sheaves, which amounts to checking:
$$
\operatorname{res}^{U, V}_{\mathcal{G}}\circ \varphi_{U} = \varphi_{V}\circ\operatorname{res}^{U, V}_{\mathcal{F}}
$$
for each $\emptyset\neq V\subseteq U\subseteq X$. This translates into checking $\varphi_{U} = \varphi_{V}\circ \operatorname{res}_{\mathcal{F}}^{U, V}$ which is clear because $\varphi_{U}=g_{U}^{-1}$, $\varphi_{V}=g_{V}^{-1}$ and $\operatorname{res}_{\mathcal{F}}^{U, V} = g_{V}\circ g_{U}^{-1}$. 
The implication (ii) $\Rightarrow$ (iii) is obvious. Next, I will think about (iii) $\Rightarrow$ (i) but I suspect that is also not hard if we use irreducibility of the space. I'd be happy to see alternative viewpoints of course!
Update: Here is an argument for (iii) $\Rightarrow$ (i): 
Suppose that $\mathcal{F}$ is locally free. Then we can find open cover ${U_{i}}$ of $X$ such that $\mathcal{F}|_{U_i}$ is the constant sheaf $B_i$. For each $i, j$, we have $U_i\cap U_{j}\neq\emptyset$ (since $X$ is irreducible), and we have a glueing map $\gamma_{ij}: B_{i} = \mathcal{F}(U_i\cap U_j) \to \mathcal{F}(U_i\cap U_j) =  B_{j}$ satisfying the cocycle condition $\gamma_{ij} = \gamma_{kj}\circ\gamma_{ik}$ on the triple intersections. In particular, $B_i$ are all isomorphic (as sets, or could be as abelian groups). Suppose that $U$ is a non-empty open set. Fix any $i$. Then $U\cap U_i\neq\emptyset$. If  $s\in U$ is a section , then $s$ is uniquely determined by its restriction $s_i\in\mathcal{F}(U\cap U_i)$. Indeed, the relation $s_j = \gamma_{ij}(s_i)$ for every other $j$ determines $s_j$. Hence, the restriction map $\mathcal{F}(U) \to \mathcal{F}(U\cap U_i) = \mathcal{F}(U_i)$ is a bijection. In particular this applies to $X$ itself. Now $\mathcal{F}(X)\to\mathcal{F}(U)\to\mathcal{F}(U_i)$ where the maps $\mathcal{F}(X)\to\mathcal{F}(U_i)$ and $\mathcal{F}(U)\to\mathcal{F}(U_i)$ are bijections. Therefore, $\mathcal{F}(X)\to\mathcal{F}(U)$ is a bijection as well.
