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Let $X$ be a set with a preorder $\le$ and $G$ be a groupoid (i.e., a category where every morphism is an isomorphism).

Treat $X$ as a category with arrows $x \to y$ whenever $x \le y$. Also, let $\sim$ be the equivalence closure of $\le$ (i.e., the smallest equivalence relation containing $\le$). One then has $x \sim y$ if and only if there are $x_0, x_1, ..., x_n$ for which $x_0=x$ and $x_n=y$ and $x_i \le x_{i+1}$ or $x_{i+1} \le x_i$ (possibly both) for $0 \le i < n$. Being an equivalence relation, $(X, \sim)$ could then also be treated as a category (in fact, a groupoid).

Then, does every functor $(X, \le) \to G$ factor uniquely through $(X, \sim)$ via the obvious functor $(X, \le) \to (X, \sim)$ (which is the identity on objects)?

It seems obvious how to do this. Namely, given any $x, y \in X$ with $x \sim y$, there are $x_0, x_1, ..., x_n$ for which $x_0=x$ and $x_n=y$ and $x_i \le x_{i+1}$ or $x_{i+1} \le x_i$ (possibly both) for $0 \le i < n$ (thus producing a "zigzag" of left and right arrows), and one could then apply the functor to each arrow in the zigzag and invert the left arrows to produce a composable chain of arrows in $G$ whose composite should then define the action of the functor on $x \sim y$.

But perhaps, that may not be well-defined, because there could be multiple "zigzags" starting with $x$ and ending with $y$.

Geoffrey Trang
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No. Every functor from a preorder $X$ to a groupoid factors through the localization of $X$ where you invert all morphisms, and the result can be a nontrivial groupoid (in the sense that it can have non-identity automorphisms), exactly because of the possibility of multiple zigzags. I think the minimal example is the "circle" $\require{AMScd}$ \begin{CD} a @>>> b \\ @VVV @AAA\\ c @<<< d \end{CD}

whose localization contains two zigzags $a \to b \leftarrow d$ and $a \to c \leftarrow d$ which are not equal and hence which define a nontrivial loop $a \to c \leftarrow d \to b \leftarrow a$.

This construction can be understood as a "suspension"; see Eric Wofsey's answer here and the picture of it here for the "$2$-sphere" version of this.

Qiaochu Yuan
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