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.

  • $\begingroup$ 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. $\endgroup$ – 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.


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