Bijective map between two definition for extension for sheaf on base

Question

Let $F$ be sheaf on base $\mathscr B =${$B_i,i \in I$}, there are two equivalent definitions for the extention for $F$, denoted by $\mathscr F $:

Def1:$\mathscr F(U) $ := {$(f_p \in F_p)_{p \in U}:$ for all $p \in U$, there exists $B \in \mathscr B$ with $p \in B \subset U,s \in F(B)$, such that $s_q = f_q$ for all $q \in B$}.

Def2:$\mathscr F(U) := \lim \limits_{\longleftarrow V \in U, V \subset \mathscr B} F(V)$ = {the set of families $(f_V)_{V \in U, V \subset \mathscr B} \in \prod_{V \in U, V \subset \mathscr B} F(V)$ such that $res_{V,W}(f_V) = f_W$ whenever $W \subset V \subset U$ with $V,W \in \mathscr B$}.

Can anyone establish a bijective map between these two definitions?

Update: Let $F_1(U)$ be def1, $F_2(U)$ be def2, construct $\phi$ : $F_2(U) \longrightarrow F_1(U)$ as follows: suppose $x = (f_{B_{i,U}})_{B_{i,U} \subset U} \in F_2(U)$, $\phi(x) = (f_p)_{p \in U}$ where if $p \in B_{i,U}$, then $f_p =$ the germ $(f_{B_{i,U}},B_{i,U})$ at point $p$.

1. $\phi(x)$ is well-defined. We have to show that if $p \in B_{i,U}$ and $p \in B_{j,U}$, the germ $(f_{B_{i,U}},B_{i,U})$ equals the germ $(f_{B_{j,U}},B_{j,U})$ at point $p$, and also $\phi(x) \in F_1(U)$. There exists at least one $B_{k,U}$ such that $p \in B_{k,U} \subset B_{i,U} \cap B_{j,U}$ since $\mathscr B$ is a base. And $x =(f_{B_{i,U}})_{B_{i,U} \subset U} \in F_2(U) (= \lim \limits_{\longleftarrow B_{i,U} \in \mathscr B} F(B_{i,U}))$, then $f_{B_{i,U}} \mid B_{k,U} = f_{B_{j,U}} \mid B_{k,U}$, which means the germ $(f_{B_{i,U}},B_{i,U})$ equals the germ $(f_{B_{j,U}},B_{j,U})$ at point $p$. And for all $p \in U$, there exists $p \in B_{i,U}$ and $f_{B_{i,U}} \in F(B_{i,U})$, such that for all $q \in B_{i,U}$,$f_q =$ the germ $(f_{B_{i,U}},B_{i,U})$ at point $q$ by the definition of $\phi$.

2. $\phi$ is injective. If $x_1 =(m_{B_{i,U}})_{B_{i,U} \subset U} \in F_2(U)$ and $x_2 =(n_{B_{i,U}})_{B_{i,U} \subset U} \in F_2(U)$ such that $\phi(x_1) = \phi(x_2)$, which means that for any $p$ and $i$ such that $p \in B_{i,U}$, the germ $(m_{B_{i,U}},B_{i,U})$ equals the germ $(n_{B_{i,U}},B_{i,U})$ at point $p$, that is $\exists p \in B_{k,U} \subset B_{i,U}$ such that $m_{B_{i,U}} \mid B_{k,U} = n_{B_{i,U}} \mid B_{k,U}$, since $p$ may go through every point in $B_{i,U}$ and $F$ is a sheaf on base which obeys the Base Identity Axiom, thus $m_{B_{i,U}} = n_{B_{i,U}}$, that is, $x_1=x_2$.

But I failed to prove that $\phi$ is subjective. Why is $\phi$ subjective?

Answer 1

Let $(f_p\in F_p)_{p\in U}\in F_1(U)$ be an element. For $p\in U$, let $B_p\in\mathscr{B}$ be a basis element such that $p\in B_p\subseteq U$ and such that there exists $f_{B_p}\in F(B_p)$ with $(f_{B_p})_q=f_q$ for all $q\in B_p$. Now take any $B\in\mathscr{B}$ such that $B\subseteq U$. As all their germs agree, the set of sections $\{\operatorname{res}_{B_p,B\cap B_p}(f_{B_p})\}_{p\in U}$ are compatible on the intersections, and thus glue to a sections $f_B$ on $B=\bigcup_{P\in U}B\cap B_p$. Again as all the germs agree, we have that if $B=B_p$ for some $p$ then $f_B=f_{B_p}$, and that $\operatorname{res}_{B,B'}(f_B)=f_{B'}$. Thus $(f_B)_{B\in\mathscr{B},\ B\subseteq U}$ is an element of $F_2(U)$, and by construction $\phi((f_B)_B)=(f_p\in F_p)_{p\in U}$.