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 ?

  • 1
    $\begingroup$ This is almost never the case, the main problem is that the system is not filtrant. You should work out $\mathcal{C}=\mathsf{Ab}$ for yourself. Think about all f.g. subgroups. $\endgroup$ – Martin Brandenburg Dec 31 '13 at 14:08
  • $\begingroup$ The family $(X_i)$ being complete under intersection, $\varinjlim X_i$ is the object $\sum A_i$ in the answer to 2. To quote Zhen Lin's answer : "it is not guaranteed to give a subobject of $A$ unless the category is nice enough". Is this false (no nice enough category exists) ? $\endgroup$ – Arkandias Dec 31 '13 at 14:38
  • 1
    $\begingroup$ No, $\varinjlim_i X_i$ is not $\sum_i X_i$. $\endgroup$ – Martin Brandenburg Jan 1 '14 at 16:58

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.


this is true if $I$ is finite because an abelian category has effective unions, see: http://www.math.mcgill.ca/barr/papers/effun.pdf

  • $\begingroup$ Actually this is false, even for $I$ finite. Take $X=\mathbb{R}^2$, $X_1=\mathbb{R}(1,0)$, $X_2=\mathbb{R}(0,1)$ and $X_3=\mathbb{R}(1,1)$. $\endgroup$ – Arkandias Nov 10 '14 at 8:01

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