# Surjectivity in exact sequence coming from the sheaf axioms

Let $\mathcal{O}_{X}$ be a sheaf of rings on a topological space $X$, and let $U\subseteq X$ be an open set. It is common to give the sheaf axioms as exactness of a sequence. For any open covering $\left\{U_{i}\right\}$ of $U$, we have the exact sequence:

$0 \to \mathcal{O}_{X}(U) \to \prod_{i}\mathcal{O}_{X}(U_{i}) \to \prod_{i,j} \mathcal{O}_{X}(U_{i}\cap U_{j})$

The last map's projection to each factor sends a sequence $(s_{i}) \mapsto s_{i}\vert_{U_{i}\cap U_{j}} - s_{j}\vert_{U_{i}\cap U_{j}}$.

My question: Are there conditions in which this last map is a surjection? Do these kinds of sheaves have a name?

Thanks!

Surjectivity at that point is a very tough restriction as we need to consider all possible choices of $\{U_i\}$. Suppose $U_i=U_j=U_k$. Then for three arbitrary elements $s_{i,j}, s_{i,k}, s_{j,k}$ of $\mathcal O_X(U_i\cap U_j)=\mathcal O_X(U_i\cap U_k)=\mathcal O_X(U_j\cap U_k)$, you'd need to exhibit $s_i,s_j,s_k\in \mathcal O_X(U_i)=\mathcal O_X(U_j)=\mathcal O_X(U_k)$ such that $s_{i,j}=s_i-s_j$, $s_{i,k}=s_i-s_k$, $s_{j,k}=s_j-s_k$. Hence we obtain the condition $s_{i,j}+s_{j,k}=s_{i,k}$, which is not in general true. The only way to make this work is to have $\mathcal O_X(U)=0$ for all $U$.