3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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... $\endgroup$ – Zhen Lin Feb 4 '16 at 13:29
  • $\begingroup$ @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? $\endgroup$ – StudentType Feb 4 '16 at 13:50
  • $\begingroup$ 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... $\endgroup$ – Zhen Lin Feb 4 '16 at 13:57
  • $\begingroup$ @ZhenLin I get it now, thanks! $\endgroup$ – StudentType Feb 4 '16 at 14:15
6
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.