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Let $\mathcal A$ be a cocomplete abelian category, let $X$ be an object of $\mathcal A$ and let $I$ be a set. Let $\{ X_i \xrightarrow{f_i} X\}_{i \in I}$ be a set of subobjects. This means we get an exact sequence $$ 0 \longrightarrow X_i \xrightarrow{f_i} X \xrightarrow{q_i}X/X_i \longrightarrow 0 $$ for each $i \in I$. It is supposed to follow (Lemma 5 in the wonderful answer to this question) that there is an exact sequence $$ \operatorname{colim} X_i \longrightarrow X \longrightarrow\operatorname{colim} X/X_i \longrightarrow 0 $$ from the fact that the colimit functor preserves colimits (and in particular, cokernels). However I do not see why this follows.

The family of exact sequences I mentioned above is equivalent to specifying the exact sequence $$ 0 \longrightarrow X_\bullet \xrightarrow{f} \Delta X \xrightarrow{q} X / X_\bullet \longrightarrow 0 $$ in the functor category $[I, \mathcal A]$, where $\Delta X$ is the constant functor sending everything to $X$. However applying the colimit functor to this sequence does not give the one we want, because the colimit of $\Delta X$ is the $I$th power of $X$ since $I$ is discrete.

Can anybody help with this? Thank you and Merry Christmas in advance!

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You probably want the colimit to be filtered (or directed) –  Mariano Suárez-Alvarez Dec 24 '12 at 18:52
@Mariano perhaps this should have been emphasised in the source. However in the source it is asserted that the other two colimits are just in fact coproducts. Is it true that we get the coproduct when we take the colimit of $ X_\bullet $and $ X / X_\bullet $ if $ I $ is directed/filtered? –  Paul Slevin Dec 24 '12 at 19:55
unless $\bigoplus $ is sometimes used as a symbol for colimit? –  Paul Slevin Dec 24 '12 at 23:18

1 Answer 1

up vote 2 down vote accepted

I think you may have misquoted the question, because if $I$ is (as you wrote) merely a set, then a colimit over it is just a direct sum.

Anyway, let me point out why "the colimit functor preserves colimits (and in particular cokernels)" is relevant. Exactness of a sequence of the form $A\to B\to C\to0$ is equivalent to saying that $B\to C$ is the cokernel of $A\to B$. So the short exact sequence you began with contains some cokernel information (plus some irrelevant information thanks to the $0$ at the left end), and what you're trying to prove is also cokernel information. The latter comes from the former by applying a colimit functor, provided colim$(X)$ is just $X$. That last proviso is why I think you've misquoted the question, since it won't be satisfied if $I$ is just a set (with 2 or more elements).

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