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I have a question regarding proof of said proposition in Borceux' book Handbook of Category Theory I (the proof is on page 80).

Let $\mathsf{I}$ be a small filtered category and $F\colon\mathsf{I}\to\mathsf{Ab}$ a functor. If $U\colon\mathsf{Ab}\to \mathsf{Set}$ is the forgetful functor, then the canonical colimit of $UF$ is $\left(\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)\right)/{\sim}$ where $\sim$ is defined as follows: $(i,x) \sim (j,y)$ if and only if there are $k \in \mathsf{Ob(I)}$ and morphisms $f\colon i\to k, g\colon j\to k$ such that $F(f)(x) = F(g)(y)$. The colimit cocone is given by functions $\lambda_i\colon F(i)\to \left(\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)\right)/{\sim}$ which map $x \in F(i)$ to $[(i,x)] \in \left(\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)\right)/{\sim}$.

To prove the proposition, one needs to define a group structure on $\left(\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)\right)/{\sim}$ for which $\lambda_i\colon F(i)\to \left(\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)\right)/{\sim}$ would be group homomorphism. Borceux defines it as follows: for $[(i,x)], [(j,y)] \in \left(\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)\right)/{\sim}$, let $k$ be an object of $\mathsf{I}$ together with morphisms $f\colon i\to k, g\colon j\to k$. Then, by the definition of $\sim$, we have $[(i,x)] = [(k,F(f)(x))]$ and $[(j,y)] = [(k,F(g)(y))]$. Set $[(i,x)] + [(j,y)] = [(k, F(f)(x) + F(g)(y))]$.

However, I'm having trouble showing that this operation is well-defined. In particular, I can't seem to prove that the resolut doesn't depend to chosen $k$: given $[(i,x)], [(j,y)] \in \left(\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)\right)/{\sim}$, let $k$ and $k'$ be objects of $\mathsf{I}$ together with morphisms $f\colon i\to k, f'\colon i\to k', g\colon j\to k, g'\colon j\to k'$. How to show that $$[(k,F(f)(x) + F(g)(y))] = [(k',F(f')(x) + F(g')(y))]?$$ By the definition of $\sim$, we need morphisms $h\colon k\to t, h'\colon k'\to t$ such that $F(h)(F(f)(x)) = F(h')(F(f')(x))$ and $F(h)(F(g)(y)) = F(h')(F(g)(y)$. I've spent some time playing around with properties of filtered categories, but still no luck.

Strangely, Borceux references another proposition, saying that, applying the proposition, it is a "straightforawrd computation" to show that the group structure is well-defined. That proposition is the one saying that every finite diagram on a filtered category has a cocone. But I just don't see how to apply it here, suspecting that there is a typo at the number he referenced.

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  • $\begingroup$ There are a few typos in the paragraph were you explain the problem. They are not really of any consequence, but you have been so precise everywhere that I thought it might be worth pointing out. The arrow $f'$ should have $k'$ as codomain. There are also a few mentions of $y'$, but they should just be $y$. $\endgroup$ Jul 19, 2020 at 12:01
  • $\begingroup$ @MarkKamsma Thanks, fixed. $\endgroup$
    – Jxt921
    Jul 19, 2020 at 15:11

1 Answer 1

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By the proposition you mention, there is a cocone in $I$ of the diagram given by $f, f', g, g'$. Having such a cocone is precisely saying that there are $h: k \to t$, $h': k' \to t$ such that $hf = h'f'$ and $hg = h'g'$. From this it easily follows by functoriality of $F$ that $$ F(h)(F(f)(x)) = F(hf)(x) = F(h'f')(x) = F(h')(F(f')(x)), $$ and similar for $g$, $g'$ and $y$.


I always find it easier to think of $I$ as a directed poset. This is justified because for every filtered category $I$, there is a directed poset $I_0$ that admits a cofinal functor $H: I_0 \to I$. So in particular, for any diagram $F: I \to \mathcal{C}$ we have $\operatorname{colim} F \cong \operatorname{colim} FH$. For references on this, see for example Theorem 1.5 in Locally Presentable and Accessible Categories by Adamek and Rosický. Or this very precise statement on the Stacks project.

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  • $\begingroup$ Good answer, thank you. $\endgroup$
    – Jxt921
    Jul 19, 2020 at 15:13

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