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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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    unless $\bigoplus $ is sometimes used as a symbol for colimit?2012-12-24

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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).