In the definition of a coherent sheaf:
1)$\mathcal{F}$ is of finite type;
2)for any open set $U$, any $n$, any $u: \mathcal {O}^n_U \rightarrow \mathcal{F}_U$ has kernal of finite type ;
Does it imply that if $(X,\mathcal{F})$, $X$ is covered by $U_i$, where each $(U_i,\mathcal{F}_{U_i})$, is coherent, then $(X,\mathcal{F})$ is coherent?
Yes. Clearly, being of finite type, is a local condition. Hence 1). The interesting part is 2). Let $\alpha : \mathcal{O}_U^n \to F$ be a homomorphism. By assumption, the kernel of $\alpha|_{U \cap U_i}$ is of finite type. This is the kernel of $\alpha$ restricted to $U_i \cap U$. Using again that being of finite type is a local condition, we see that the kernel of $\alpha$ is of finite type.
