# Filtered colimits in the category of sets: an equivalence relation

A category $$\mathsf{I}$$ is filtered if

• it is nonempty,

• for any $$i,j \in \mathsf{I}$$ there is $$k \in \mathsf{I}$$ and morphisms $$f\colon i\to k$$ and $$g\colon j\to k$$,

• for any pair $$f,g\colon i\to j$$ of parallel morphisms in $$\mathsf{I}$$ there is $$k \in \mathsf{I}$$ and a morphism $$h\colon j\to k$$ so that $$h\circ f = h\circ g$$;

equivalently, $$\mathsf{I}$$ is filtered if for any finite diagram in $$\mathsf{I}$$ admits a cocone.

Let $$\mathsf{I}$$ be a small filtered category and let $$F\colon\mathsf{I}\to\mathsf{Set}$$ be a functor.

I want to prove that if $$\mathsf{I}$$ is filtered, then the relation $$\sim$$ on $$\bigsqcup_{i \in \mathsf{Ob(I)}} F(i)$$ so that $$(i,x) \sim (j,y)$$ precisely when there is an object $$k$$ of $$\mathsf{I}$$ together morphisms $$f\colon i\to k$$ and $$g\colon j\to k$$ so that $$F(f)(x) = F(g)(y)$$ is an equivalence relation.

In particular, I don't know how to prove that it is a transitive relation.

Borceux in his book "Handbook of Categorical Algebra I" attempts to prove it this way:

If $$(i,x) \sim (j,y)$$ and $$(j,y) \sim (k,z)$$, there are $$k',k'' \in \mathsf{I}$$ together with morphisms $$f\colon i\to k', g\colon j\to k', f'\colon j\to k''$$ and $$g'\colon k\to k''$$ so that $$F(f)(x) = F(g)(y)$$ and $$F(f')(y) = F(g')(z)$$. We can consider a finite category $$\mathsf{J}$$ consisting of objects $$i,j,k,k',k''$$ and morphisms $$f,g,f',g'$$ (aside from identity morphisms) and diagram $$D\colon \mathsf{J}\to\mathsf{I}$$ which the must have a cocone.

I'm having trouble with his construction of the finite category $$\mathsf{J}$$, as it seems to implicitly assume that neither pair of $$f,g,f',g'$$ is composable in $$\mathsf{I}$$. However, what if, for example, we have $$k' = j$$? Or $$k = k''$$? The category $$\mathsf{J}$$ then would have to contain the compositions as well. If we define it as a subcategory of $$\mathsf{I}$$ containing said objects and morphisms, then it doesn't have to finite.

Let $$(i_1,x_1),(i_2,x_2),(i_3,x_3)\in\coprod_{i\in Obj(I)}F(i)$$, such that $$(i_1,x_1)\sim(i_2,x_2)$$ and $$(i_2,x_2)\sim(i_3,x_3)$$. Then by the definition of this relation there exist such $$(i_4,x_4),(i_5,x_5)\in\coprod_{i\in Obj(I)}F(i)$$ and morphisms $$f\colon i_1\to i_4$$, $$g\colon i_2\to i_4$$, $$h\colon i_2\to i_5$$, $$k\colon i_3\to i_5$$, that $$(F(f))(x_1)=x_4$$, $$(F(g))(x_2)=x_4$$, $$(F(h))(x_2)=x_5$$, $$(F(k))(x_3)=x_5$$. By filteredness, take $$i_6\in I$$, such that there are some morphisms $$n\colon i_4\to i_6$$ and $$m\colon i_5\to i_6$$, and then take $$i_7\in I$$, such that there is a morphism $$l\colon i_6\to i_7$$, such that $$l\circ n\circ g=l\circ m\circ h$$. Then consider the pair $$(i_7,F(l\circ n\circ f)(x_1))$$. Note, that: $$(F(l\circ m\circ k))(x_3)=(F(l)\circ F(m)\circ F(k))(x_3)=(F(l)\circ F(m))((F(k))(x_3))=$$ $$=(F(l)\circ F(m))((F(h))(x_2))=(F(l)\circ F(m)\circ F(h))(x_2)=(F(l\circ m\circ h))(x_2)=$$ $$=(F(l\circ n\circ g))(x_2)=(F(l)\circ F(n)\circ F(g))(x_2)=(F(l)\circ F(n))((F(g))(x_2))=$$ $$(F(l)\circ F(n))((F(f))(x_1))=(F(l)\circ F(n)\circ F(f))(x_1)=(F(l\circ n\circ f))(x_1),$$ what proves transitivity (it is a straighforward proof).
As for cocones: objects and morphisms of $$J$$ should obviously be disjoint, so objects of $$J$$ are not $$i,j,k,k',k''$$, but, for example, $$0,1,2,3,4$$ and $$D(0)=i$$, $$D(1)=j$$ and so on.
• Of course! I'm almost ashamed I didn't think of taking the category with objects $0,1,2,3,4$ and morphisms $0 \to 3, 1\to 3, 1\to 4, 2\to 4$ myself. Thanks, Oskar. – Jxt921 Jan 22 at 9:47
There’s a functor from $$I$$ no matter what. For somesimpler examples, there’s a functor from the non-commutative square to the commutative square, and there’s a functor from the natural numbers, seen as a poset, to the natural numbers, seen as a monoid. Better yet, every category maps to the terminal category, where everything is identified. It never hurts to add more relations in the codomain of a functor.