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At least if we have a Grothendieck category everything seems OK: Suppose $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ is exact with $M'$ and $M''$ of finite type. Assume $\{M_i\}$ is a directed collection of subobjects of $M$ such that $\sum_i M_i=M$. We then have $M'=M'\bigcap\sum_i M_i=\sum_i M'\bigcap M_i$ and hence $M'=M'\bigcap M_{i_0}$ for some $i_0$. Throwing away all indices which are not $\geq i_0$ we may assume $M'\subseteq M_i$ for all $i$. We then get that $M''=\sum_i M_i/M'$ and hence $M''=M_{i_1}/M'$ for some $i_1$ which means that $M=M_{i_1}$.