# Adjoint of the inclusion functor from Preord to Cat

Suppose $$F : Preord \to Cat$$ is the inclusion functor. Suppose that $$G : Cat \to Preord$$ is the functor which maps each category $$C$$ to its associated preorder (each object is an element and $$X \leq Y$$ iff there is a morphism from $$X$$ to $$Y$$ in $$C$$).

To show that $$G$$ is the left adjoint to $$F$$, I'm trying to come up with an isomorphism $$\phi : hom(GC, D) \to hom(C, FD)$$ but I can't seem to make sense of what it should look like.

Warning: I'm not exhibiting the mapping $$\phi$$ per se, I'm just giving another way to prove that $$G$$ is the left adjoint.

Here is a useful lemma:

Lemma. A functor $$F: \mathbf D\to \mathbf C$$ has a left adjoint if and only if each comma category $$(c \downarrow F)$$ admits a initial object $$(d_c, a_c : c \to F(d_c))$$. In that case the left adjoint $$G$$ is given by $$c \mapsto d_c \quad , \quad (c\overset f \to {c'}) \mapsto (d_c \overset {G(f)} \to d_{c'})$$ where $$G(f)$$ is the map induced by the initiality of $$a_c$$ applied on $$a_c'\circ f$$.

So in your case, given a small category $$c$$, let us write $$a_c : c \to \bar c$$ for the functor that collapses every parallel arrows, so that $$\bar c$$ is the preorder associated to $$c$$ viewed as a category (that is $$F(G(c))$$ in your notation). You now have to check that:

• each functor $$c \to p$$ from a small category $$c$$ into a preorder $$p$$ can be factored in a unique way through the functor $$a_c$$ (mapping a category to a preorder only cares about the action on the objects, the action on the arrows is coerced after that),
• for each functor $$f : c \to c'$$ such a factorization applied to the composite $$a_{c'} \circ f$$ yields indeed the functor $$F(G(f))$$ (this is kind of immediate).

This is assuming $$Cat$$ is the category of small categories, otherwise $$GC$$ (for $$C$$ in $$Cat$$) would not be a set and can thus not be a preorder.

You can define the map $$\phi: \hom(GC, D) \to \hom(C, FD)$$ as follows. Given a morphism of preorders $$f: GC \to D$$, we define a functor $$g: C \to FD$$ by letting $$g$$ be the same as $$f$$ on objects (since the objects in $$C$$ and $$FD$$ are the elements in $$GC$$ and $$D$$ respectively). For any arrow $$X \to Y$$ in $$C$$, we have that $$X \leq Y$$ in $$GC$$, so since $$f$$ is a morphism of preorders we have $$f(X) \leq f(Y)$$ in $$D$$. That means that we have exactly one arrow $$g(X) \to g(Y)$$ in $$FD$$, so we send the arrow $$X \to Y$$ there. It should be clear that this defines an actual functor $$g: C \to FD$$.

The operation $$\phi$$ has an inverse, by taking a functor $$g: C \to FD$$ and defining a morphism of preorders $$f: GC \to D$$ by letting $$f$$ be setting $$f(X) = g(X)$$ for all elements in $$GC$$ (again, which were objects in $$C$$). The fact that $$g$$ was a functor now guarantees that $$f$$ is a morphism of preorders.

Essentially this is saying that there is a unique way to extend a morphism of preorders to a functor of categories, because there is exactly one place we can send the arrows.

Of course, you would still have to check that $$\phi$$ is a natural bijection, but that is not hard and I leave that as an exercise to the reader.