Let $ \mathsf I $ be a category. Let's call $ \mathsf I $ filtered if
- $ \mathsf I $ has at least one object;
- for every $ i,j\in \mathsf I $, there exists $ k\in \mathsf I $ and morphisms $ i\to k $, $ j\to k $;
- for every $ i,j\in \mathsf I $ and for every pair of parallel morphism $ \sigma,\tau\colon i\rightrightarrows j $, there exists $ k\in \mathsf I $ and a morphism $ \pi \colon j\to k $ such that $ \pi \circ \sigma = \pi\circ \tau $.
I'm trying to show the following.
Let $ \mathsf I $ be a category. Then $ \mathsf I $ is filtered if and only if every finite diagram in $ \mathsf I $ has a cocone.
Unfortunately, I got stuck in proving that $ \mathsf I $ filtered implies the existence of cocones. Here's what I tried so far.
Let $ \mathsf I $ be filtered, and let $ F\colon \mathsf C\to \mathsf I $ be a functor from a finite category $ \mathsf C $ to $ \mathsf I $. If $ \mathsf C $ has only one object $ X_1 $, then by (2) there is $ i\in \mathsf I $ and an arrow $ \colon FX_1\to i $. We call $ \mu_{X_1} $ this arrow $ \mu_{X_1}\colon Fx_1\to i $. If $ \mathsf C $ has two objects $ X_1 $ and $ X_2 $, then again by (2) there are $ i_1,i_2\in \mathsf I $ and arrows $ FX\to i_1 $, $ FY\to i_2 $, and applying (2) another time we obtain $ k\in \mathsf I $ and two arrows $ i_1\to k $, $ i_2\to k $. This time we call $ \mu_{X_1} $ and $ \mu_{X_2} $ the arrows $ \mu_{X_1},\mu_{X_2}\colon FX_1,FX_2\to k $ obtained by composing the arrows $ FX_r\to i_r $ and $ i_r\to k $, for $ r = 1,2 $. By induction, if $ \mathsf C $ has $ n $ elements then there exists $ k\in \mathsf I $ and a family $ \mu = {\left(\mu_{X}\colon FX\to k\right)}_{X\in \mathsf C} $ of morphisms $ \mu_X\colon FX\to k $, one for every $ X\in \mathsf C $. I've not been able to go further, and show that the family $ \mu $ satisfies the cocone condition. How can I prove that for every $ f\colon X\to Y $ in $ \mathsf C $, $ \mu_Y\circ Ff = \mu_X $?