Adjoint of the inclusion functor from Preord to Cat

Question

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.

Answer 1

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).

Answer 2

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.