# Relationship between functors

You will have to forgive me as I am very new to category theory - fifth of the way through Categories for a working mathematician. I'm interested in the following;

Let $F:A \to B$ and $G:A \to C$ be full functors, where $A$,$B$ and $C$ are groupoids and $B$ has a single object. Let $G$ be such that $Gf=Gf'$ if and only if $Ff=Ff'$ where $f$ and $f'$ are morphisms of $A$. What is the relationship of $G$ to $F$?

If $C$ is also a category with a single object then $C$ and $B$ are isomorphic?

2
2022-07-25 20:46:33
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Surprisingly, even under all the hypotheses, the answer to your first question is essentially negative.$\DeclareMathOperator{\ob}{ob}$ $\DeclareMathOperator{\Hom}{Hom}$Let $\mathbb{A}$ be two disjoint copies of the infinite cyclic group $\mathbb{Z}$, let $\mathbb{B}$ be a single copy of $\mathbb{Z}$, and let $\mathbb{C}$ be a connected pair of copies of $\mathbb{Z}$. More precisely:

• $\ob \mathbb{A} = \{ A_1, A_2 \}$, $\Hom(A_1, A_1) = \mathbb{Z}$, $\Hom(A_2, A_2) = \mathbb{Z}$, $\Hom(A_1, A_2) = \emptyset$, $\Hom(A_2, A_1) = \emptyset$.

• $\ob \mathbb{B} = \{ B \}$, $\Hom(B, B) = \mathbb{Z}$.

• $\operatorname{ob} \mathbb{C} = \{ C_1, C_2 \}$, $\Hom(C_1, C_1) = \mathbb{Z}$, $\Hom(C_2, C_2) = \mathbb{Z}$, $\Hom(C_1, C_2) = \mathbb{Z}$, $\Hom(C_2, C_1) = \mathbb{Z}$, composition given by addition.

Now, there is a "bad" functor $F : \mathbb{A} \to \mathbb{B}$ constructed as follows: on $A_1$, it acts as the identity, but on $A_2$, it acts as the outer automorphism $x \mapsto -x$. Let $G : \mathbb{A} \to \mathbb{C}$ be the obvious functor that acts as the identity on morphisms. I claim there is no functor $H : \mathbb{C} \to \mathbb{B}$ such that $H G \cong F$. Indeed, if $H$ were such a functor, then there would have to be an integer $n$ such that $$n + x = -x + n$$ for all $x \in \mathbb{Z}$, which is patent nonsense. (Basically, two points in the same connected component of a groupoid can only be related by an inner automorphism.)

Happily, the answer to your second question is affirmative. Something slightly stronger is true:

Proposition. If $\mathbb{A}$ is category, $\mathbb{B}$ and $\mathbb{C}$ are both categories with one object, $F : \mathbb{A} \to \mathbb{B}$ and $G : \mathbb{A} \to \mathbb{C}$ are full functors, and $F f = F f'$ if and only if $G f = G f'$, then there is a unique functor $H : \mathbb{C} \to \mathbb{B}$ such that $H G = F$ and $H$ is an isomorphism.

The proof is obvious and has more to do with algebra than category theory. As an additional exercise, you might want to try to explain why this can be done without the axiom of choice.

1
2022-07-25 21:22:16
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