An object $M$ of an abelian category is called of finite type iff for every directed set of subobjects $M\_i$ of $M$ whose sum is $M$ there exists some $i$ with $M = M\_i$. Is the direct sum $M \oplus N$ of two objects $M,N$ of finite type again of finite type?


So let $P\_i \subseteq M \oplus N$ be a directed set of subobjects whose sum is $M \oplus N$. If $M\_i$ denotes the projection of $P\_i$ to $M$ and $N\_i$ the one to $N$, then it is easy to see that there is some $i$ with $M\_i = M$ and $N = N\_i$. But does not show yet $P\_i = M \oplus N$!


More general: If $0 \to M' \to M \to M'' \to 0$ is an exact sequence, where $M',M''$ are of finite type, does this imply that $M$ is of finite type?


If there are counterexamples: Is it at least true in a Grothendieck-category? What are other reasonable definitions for "finitely generated" which generalize the ones for modules over rings or quasi-coherent modules over nice schemes?



$M \oplus N$ is of finite type if $M,N$ are of finite type?
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
$i\_1$


**Addendum**: Stealing some ideas from Sándor's reply we can get the
statement without extra axioms. Note that finite generation is formulated in
terms of $\sum\_iM\_i=M$ which is the same as $\mathrm{lim}M\_i\to M$ being surjective
(as the sum is image of the limit). Now, with notations as before we put $M''\_i$
to be the image of $M\_i$ in $M''$. As $\mathrm{lim}M\_i\to M$ is surjective we get
that so is $\mathrm{lim}M\_i''\to M''$ and hence $M''\_i=M''$ for some $i$ and after
throwing away we can assume this is always true. This means that we get an exact
sequence $0\to M'\_i\to M'\to M/M\_i\to0$ and as $\mathrm{lim}M/M\_i=0$ (by right exactness of
directed colimits) we get that $\mathrm{lim}M'\_i\to M'$ is surjective (again by
right exactness) and hence that $M'\_i=M'$ for some $i$ but then $M\_i=M$ as $M''\_i=M''$,
 which means that $M=M\_{i}$.


