In an Abelian category $\mathcal{A}$, let {$A_i$} be a family of subobjects of an object $A$. How to show that if $\mathcal{A}$ is cocomplete(i.e. the coproduct always exists in $\mathcal{A}$), then there is a smallest subobject $\sum{A_i}$ of $A$ containing all of $A_i$?

Surely this $\sum{A_i}$ cannot be the coproduct of {$A_i$}, but I have no clue what it should be.

  • $\begingroup$ push out of pull back of $A_i \to A$ $\endgroup$ – Alexander Thumm Aug 20 '11 at 9:31
  • $\begingroup$ @Alexander: That only works for a pair of subobjects. If you have three subobjects, the pullback may be trivial even when there are non-trivial pairwise intersections. $\endgroup$ – Zhen Lin Aug 20 '11 at 10:01
  • $\begingroup$ @Zhen: for three objects $A_1 + A_2 + A_3 = (A_1 + A_2) + A_3$ obviously $\endgroup$ – Alexander Thumm Aug 20 '11 at 10:38
  • $\begingroup$ You seem to have found the right generalization though. $\endgroup$ – Alexander Thumm Aug 20 '11 at 10:47

You are quite right that it can't be the coproduct, since that is in general not a subobject of $A$. Here are two ways of constructing the desired subobject:

  1. As Pierre-Yves suggested in the comments, the easiest way is to take the image of the canonical map $\bigoplus_i A_i \to A$. This works in any cocomplete category with unique epi-mono factorisation.

  2. Alternatively, the subobject $\sum A_i$ can be constructed by taking the colimit over the semilattice of the $A_i$ and their intersections. This construction can be carried out in any bicomplete category, but is not guaranteed to give a subobject of $A$ unless the category is nice enough.

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    $\begingroup$ Dear Zhen Li: I'll tell you what I'd have answered, and you (or someone else) will tell me why it's incorrect, if you're patient enough. The coproduct maps naturally to $A$, with some kernel. Thus, the quotient of the coproduct by the kernel can be viewed as a subobject. $\endgroup$ – Pierre-Yves Gaillard Aug 20 '11 at 9:45
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    $\begingroup$ @Pierre-Yves: That works too. Or we can just directly ask for the image, since every morphism in an abelian category has a unique epi-mono factorisation. $\endgroup$ – Zhen Lin Aug 20 '11 at 9:45
  • $\begingroup$ Dear Zhen Lin, thank you very much! $\endgroup$ – Pierre-Yves Gaillard Aug 20 '11 at 9:47
  • $\begingroup$ +1. Nice answer! [Thanks for the edit! Sorry I had forgotten to upvote, and now it looks like I did it because you were kind enough to put my name. I hope people will think I'm self-centered, but not to that extent... ;)] $\endgroup$ – Pierre-Yves Gaillard Aug 20 '11 at 10:11
  • $\begingroup$ @Zhen Lin Thank you very much! $\endgroup$ – Li Zhan Aug 20 '11 at 10:28

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