# “Coforgetful” functors?

Let $F : \mathbf{Sets} \to \mathbf{Cat}$ be the free functor that takes the elements of a set to the objects of a discrete category. Does it has a left adjoint? Awodey (2010 p.249 ex.8) says it does, and suggests we can construct it considering the “connected components” of a category. I am not sure what this means though.

I think the forgetful functor $U : \mathbf{Cat} \to \mathbf{Sets}$ that takes a category $\mathbf{C}$ and gives us the set of its objects would work as a left adjoint (i.e. not only the usual $F \dashv U$, but also $U \dashv F$), since any functor $X \to FS$ (which is just a mapping of objects) uniquely determines a function $UX \to S$ (a mapping of its elements) and vice-versa - where $X \in \mathbf{Cat}$ and $S \in \mathbf{Sets}$ (of course the bijection of Hom-sets must be natural in $X$ and $S$ too, but I believe we can take this for granted.)

Is this correct?

• It is not correct. A functor from a category to a set is just a mapping of objects, but the morphisms still have to go somewhere... – Zhen Lin Feb 4 '16 at 13:29
• @ZhenLin mm.. I am not sure if I understand what you mean. How can I map each morphism of $X$ to a morphism of $FS$, if it has no morphisms? Could you be more is specific about where is the flaw? – StudentType Feb 4 '16 at 13:50
• If you don't send the morphisms somewhere then you don't have a functor. So you have to send the morphisms somewhere. I wouldn't say $F S$ has no morphisms – identity morphisms are still morphisms... – Zhen Lin Feb 4 '16 at 13:57
• @ZhenLin I get it now, thanks! – StudentType Feb 4 '16 at 14:15

## 1 Answer

No, this isn't correct. For a simple counterexample, consider the category $\mathsf{C} = \{ A \to B \}$, i.e. a category with two objects $A$ and $B$ and one nonidentity morphism from $A$ to $B$. Let also $X = \{a,b\}$ be a two-element set.

Then $\operatorname{Mor}_{\mathsf{Cat}}(\mathsf{C}, F(X))$ has two elements, the two constant functors $A,B \mapsto a$ and $A,B \mapsto b$. It's not possible to map $A$ and $B$ to different elements of $X$, because then the morphism $A \to B$ wouldn't have anywhere to go (there is no morphism $a \to b$ or $b \to a$ in $F(X)$. But $\operatorname{Mor}_{\mathsf{Set}}(U(\mathsf{C}), X)$ has four elements, corresponding to the four maps $\operatorname{ob}(\mathsf{C}) \to X$.

The correct functor to consider is the functor of connected components, $\operatorname{conn} : \mathsf{Cat} \to \mathsf{Set}$. Two objects $X$ and $Y$ of a category $\mathsf{C}$ are said to be in the same connected component if there exists a zigzag of morphisms $X \gets X_1 \to X_2 \gets \dots \to X_n \gets Y$. This is an equivalence relation on $\operatorname{ob} \mathsf{C}$, and the functor $\operatorname{conn}$ maps $\mathsf{C}$ to the quotient $\operatorname{conn} \mathsf{C} = \operatorname{ob} \mathsf{C} / \sim$ by this equivalence relation.

A functor $F : \mathsf{C} \to \mathsf{D}$ maps a zigzag of morphisms to a zigzag of morphisms, so the map $\operatorname{ob} \mathsf{C} \to \operatorname{ob} \mathsf{D}$ is compatible with the two equivalence relations and induces a map $\operatorname{conn} \mathsf{C} \to \operatorname{conn} \mathsf{D}$. You thus get a well-defined functor $\operatorname{conn} : \mathsf{Cat} \to \mathsf{Set}$.

Now it's an exercise to show that $\operatorname{conn} \dashv F$:

• Given a map $f : \operatorname{conn} \mathsf{C} \to X$, you get a well-defined morphism by sending all the elements of a connected component $\gamma$ to the object $f(\gamma) \in X = \operatorname{ob} F(X)$, and all the morphisms in this connected component to the identity of $f(\gamma)$. Since there are no morphisms between two different connected components, this is well-defined.

• In the other direction, a functor $\mathsf{C} \to F(X)$ has to be constant on connected components (essentially the same proof that a continuous map from a path-connected space to a discrete space is constant), and so induces a well-defined map $\operatorname{conn} \mathsf{C} \to X$.

These two constructions are inverse to each other, and thus:

$$\operatorname{Mor}_{\mathsf{Cat}}(\mathsf{C}, F(X)) \cong \operatorname{Mor}_{\mathsf{Set}}(U(\mathsf{C}), X).$$