Characterisation of abelian categories in which colimit of subobjects are subobjects This question is related to 1 and 2.
Given an abelian category $\mathcal{C}$ in which colimit exists. What is a necessary and sufficient condition on $\mathcal{C}$ so that given any $X \in \mathcal{C}$ and a family $(X_i)_{i \in I}$ of subobjects of $X$ complete under intersection, the natural morphism
$$ \varinjlim_{i \in I} X_i \to X$$
is injective (the transition maps of the colimit being the inclusions, thus not filtrant in general) ?
What about the same statement with $I$ being finite ?
I am particularily interested in Grothendieck categories or categories of modules of finite type over a noetherian ring. Does the property holds in these ?
 A: Let $\mathcal{C}$ be an abelian category, $X$ an object of $\mathcal{C}$, $(X_i)_{i \in I}$ a (possibly infinite) family of subobjects of $X$ and assume that :


*

*$(X_i)_{i \in I}$ is complete under finite intersections

*$X_i \cap \left( \sum_{j \in J} X_j \right) = \sum_{j \in J} (X_i \cap X_j)$ for all $i \in I$ and $J \subset I$ finite
Then the colimit of $(X_i)_{i \in I}$ with transition maps being inclusions exists and the natural map
$$\varinjlim_{i \in I} X_i \to \sum_{i \in I} X_i$$
is an isomorphism.
Proof. Let $(f_i : X_i \to Y)_{i \in I}$ be a compatible system of morphisms in $\mathcal{C}$. We will check that there exists a unique map $f : \sum_{i \in I} X_i \to Y$ in $\mathcal{C}$ satisfying the universal property of colimit. Uniqueness is immediate, since one must have $f(\sum_{j \in J} x_j) = \sum_{j \in J} f_j(x_j)$ for all $J \subset I$ finite and for all $(x_j)_{j \in J} \in (X_j)_{j \in J}$. In order to prove existence, we must verify that $f$ is well defined this way. Thus, we must check that any relation $\sum_{j \in J} x_j = 0$ in $X$ with $J \subset I$ finite implies that $\sum_{j \in J} f_j(x_j) = 0$ in $Y$. We proceed by induction on $|J|$ (the case $|J|=0$ is trivial). Let $i \in J$. Then $x_i = -\sum_{j \in J\backslash \{i\}} x_j$, thus $x_i \in X_i \cap (\sum_{j \in J\backslash\{i\}} X_j) = \sum_{j \in J\backslash\{i\}} (X_i \cap X_j)$, so $x_i = -\sum_{j \in J\backslash\{i\}} x_{i,j}$ with $x_{i,j} \in X_i \cap X_j$. The morphisms $(f_j)_{j \in J}$ being compatible, we have $f_i(x_i) = \sum_{j \in J\backslash\{i\}} f_j(x_{i,j})$, hence $\sum_{j \in J} f_j(x_j) = \sum_{j \in J\backslash\{i\}} f_j(x_j-x_{i,j})=0$ by induction.
A: this is true if $I$ is finite because an abelian category has effective unions, see:
http://www.math.mcgill.ca/barr/papers/effun.pdf
