I was reading this post
I am trying to show that $S^{-1}R=\operatorname{colim}F(s)$, where $S$ is a multiplicative closed set in a commutative ring $R$ and $F$ is a functor from a filtered category $I$ to mod-$R$ and $I$ is constructed as follows: objects of $I$ are just the elements of $S$ and the morphisms are $$\operatorname{Hom}_{I}(s_{1},s_{2})=\{s \in S : s_{1}s=s_{2}\}$$
By the axioms of being filtered, suppose exists $u,v \in S$, such that $su=sv$, then there must exists $t \in S$, such taht $ut=vt$?
How is this satisfied?
Let $s,t\in I$, that is $s,t\in S$. Then $t$ is a morphism $s\to st$ and $s$ a morphism $t\to st$.
Now let $s_1,s_2\in I$, $s,t$ two morphism $s_1\to s_2$. That is $s_2 = s_1s = s_1t$. Now we're looking for an element $u\in S$ such that $su = tu$. But that element is simply $s_1$ ! So $s_1$ is a morphism $s_2\to s_1s_2$ and $s_1t = s_2 = s_1s$.