Let $\mathbb I$ be a small category (i.e., its class of arrows is a set) which satisfies the following: (1) it is nonempty, (2) for each $i,i'\in \mathbb I$ there exists $i''\in \mathbb I$ and $\mathbb I$-arrows $f\colon i\to i''$ and $f'\colon i'\to i''$, and (3) for each pair of $\mathbb I$-arrows $f_0, f_1\colon i_0\to i_1$ there is a $i_2\in \mathbb I$ and an $\mathbb I$-arrow $g\colon i_1\to i_2$ such that $gf_0=gf_1$. Then we say that $\mathbb I$ is filtered.

I am attempting to prove the following equivalence:

$\mathbb I$ is filtered iff for each finite category $\mathbb J$ and diagram $D\colon \mathbb I\times\mathbb J\to \mathbf{Set}$ the natural map $\delta\colon colim_{\mathbb I}lim_{\mathbb J}(D)\to lim_{\mathbb J}colim_{\mathbb I}(D)$ is bijective.

I was able to prove the forward direction. For the converse, I was able to prove the first two conditions (1) and (2) hold. However, I am having trouble showing (3) holds. My idea is to set $\mathbb J$ to be the category with two objects and two parallel arrows $a,b\colon 0\to 1$. Then given $f_0,f_1\colon i_0\to i_1$ we define the functor

$$ D\colon \mathbb I\times \mathbb J^{op}\to \mathbf{Set},\quad (i,j)\mapsto \mathbb I(i_j,i) $$ with the obvious precomposition action on the $\mathbb J$-arrows. Then it seems that the limit over $\mathbb J$ will define a non initial functor $\mathbb I\to \mathbf{Set}$ iff the condition (3) is satisfied. However, I am having trouble how to show that delta is not just the map $\emptyset \to \emptyset$.

  • $\begingroup$ How does $D(i,j)$ behave on objects other than $i=i_0$ and $i=i_1$? $\endgroup$
    – Thorsten
    Jan 13, 2015 at 4:03
  • $\begingroup$ @Thorsten It is already defined for each $i\in\mathbb I$. It takes $(i,j)$ to the homset $\mathbb I(i_j,i)$, the set of $\mathbb I$-arrows $i_j\to i$. In other words, it is determined by the two representable $\mathbb I(i_0,\square)$ and $\mathbb I(i_1,\square)$. $\endgroup$ Jan 13, 2015 at 4:09
  • 1
    $\begingroup$ Ah you are right, I'm sorry for this! :) $\endgroup$
    – Thorsten
    Jan 13, 2015 at 4:12

1 Answer 1


You've got the right setup. The trick is that $\lim_{j \in J} \operatorname{colim}_{i \in I} I(Fj,i) = \lim_{j \in J} 1 = 1$ for any categories $I,J$, and functor $F: J^\mathrm{op} \to I$. The first equation is because the colimit of a representable functor is always 1 (exercise!).

On the other hand, $\operatorname{colim}_{i \in I} \lim_{j \in J} I(Fj,i) = \pi_0 \mathsf{Elts}(\lim_{j \in J} I(Fj,i))$ where $\mathsf{Elts}$ is the construction sending a presheaf to its category of elements and $\pi_0$ takes the connected components of this category. Now $\lim_{j \in J} I(Fj,i)$ is the functor sending $i$ to the set of cocones on $F$ with vertex $i$, and $\mathsf{Elts}(\lim_{j \in J} I(Fj,i)$ is in fact the category of cocones on $F$.

So commutation of the limit and colimit of the particular functor $I(F,1)$ is equivalent to the category of cocones on $F$ being connected. In particular, this category of cocones is nonempty as desired.

  • $\begingroup$ Sorry, an extra "op" worked its way into my $J$. $\endgroup$
    – tcamps
    Jan 16, 2015 at 3:12
  • $\begingroup$ thanks! the fact that colimits of representables is 1 seems quite useful $\endgroup$ Jan 16, 2015 at 4:04

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