# Forgetful functor $\mathsf{Ab}\to \mathsf{Set}$ preserves filtered colimits: the group structure on set-theoretic filtered colimit

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.

• 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 $y$. Jul 19, 2020 at 12:01
• @MarkKamsma Thanks, fixed. Jul 19, 2020 at 15:11

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.