The last property that makes a presheaf a sheaf says: Let $U$ be an open set in $X$ such that $U=\bigcup_{j\in J}U_j$.
If $aj \in \mathscr{F}(U_j)$, $j \in J$, satisfies
$\rho_{U_j(U_i\cap U_j)}(a_j)=\rho_{U_i(U_j\cap U_i)}(a_i)$, $i,j \in J$,
then there exists $a \in \mathscr{F}(U)$ such that $a_j=\rho_{UU_j}(a)$.
My question is as follows, $\forall j\in J, \exists a_j \in \mathscr{F}(U_j)$ such that, $\forall i,j \in J, \rho_{U_j(U_i\cap U_j)}(a_j)=\rho_{U_i(U_j\cap U_i)}(a_i)$, then there...
How can there be both variations on $j$ at the same time? It seems to me that the second variation of $j$ (together with the variation of $i$) in $J$ doesn't happen at all.