Options

Question 2.1 - Functors out of Set

\(\require{begingroup}\begingroup\require{AMScd}\newcommand{\tlap}[1]{\raisebox{0pt}[0pt][0pt]{#1}}\newcommand{\blap}[1]{\vbox to 0pt{\hbox{#1}\vss}}\) Functors out of Set.

Consider the category 3, which has three objects and six morphisms, and is depicted as follows:

\begin{CD} 1 @>a>> 2 @>b>>3 \end{CD}

How many functors are there from Set to 3? Write them down.


Next All

Comments

  • 1.
    edited June 14

    I'll start the ball rolling by showing that the answer is least three and at most nine.

    For starters, there are the three constant functors: one which maps everything to 1, another to 2, and another to 3.

    Proposition. Let \(F\) be a functor from \(Set\) to \(3\). If \(A\) and \(B\) are non-empty sets, then \(F(A) = F(B)\).

    Proof. Since \(A\) and \(B\) are non empty, there are functions going from \(A\) to \(B\), and from \(B\) to \(A\). For a contradiction, suppose \(F(A) = m\), \(F(B) = n\), and \(m < n\). Let \(g: B \rightarrow A\) be some function in Set. Then \(F\) must map \(g\) to some morphism \(F(g): n \rightarrow m\). However, because \(m < n\), no such arrow exists.

    So, let \(Set_1 = Set - \lbrace\rbrace\), all sets except the empty set.

    The proposition says that everything in \(Set_1\) must map to the same object under \(F\). The behavior of \(F\) is therefore fully characterized by specifying which object \(Set_1\) goes to, and which object the empty set goes to.

    There are three choices for each of these. Hence there are nine candidate functors.

    However, there are further constraints, so the answer looks to be less than nine.

    That is the sound of the ball rolling...

    Comment Source:I'll start the ball rolling by showing that the answer is least three and at most nine. For starters, there are the three constant functors: one which maps everything to 1, another to 2, and another to 3. **Proposition.** Let \\(F\\) be a functor from \\(Set\\) to \\(3\\). If \\(A\\) and \\(B\\) are non-empty sets, then \\(F(A) = F(B)\\). **Proof.** Since \\(A\\) and \\(B\\) are non empty, there are functions going from \\(A\\) to \\(B\\), and from \\(B\\) to \\(A\\). For a contradiction, suppose \\(F(A) = m\\), \\(F(B) = n\\), and \\(m < n\\). Let \\(g: B \rightarrow A\\) be some function in Set. Then \\(F\\) must map \\(g\\) to some morphism \\(F(g): n \rightarrow m\\). However, because \\(m < n\\), no such arrow exists. So, let \\(Set_1 = Set - \lbrace\rbrace\\), all sets except the empty set. The proposition says that everything in \\(Set_1\\) must map to the same object under \\(F\\). The behavior of \\(F\\) is therefore fully characterized by specifying which object \\(Set_1\\) goes to, and which object the empty set goes to. There are three choices for each of these. Hence there are nine candidate functors. However, there are further constraints, so the answer looks to be less than nine. That is the sound of the ball rolling...
  • 2.
    edited January 20

    Sub-exercise. Someone could raise the following objection to my argument above. They could say that I have only talked about the object component of the functor. When we take into account the behavior of the functor on morphisms as well, there could in principle be many many more functors.

    The sub-exercise is to add a brief postscript to the above argument, showing why, in this particular case, the morphisms don't end up affecting the counting logic that I gave above.

    Comment Source:Sub-exercise. Someone could raise the following objection to my argument above. They could say that I have only talked about the object component of the functor. When we take into account the behavior of the functor on morphisms as well, there could in principle be many many more functors. The sub-exercise is to add a brief postscript to the above argument, showing why, in this particular case, the morphisms don't end up affecting the counting logic that I gave above.
  • 3.
    edited January 21

    since there's always a function from the empty set to any set but not the other way around, we are down to a total of 6 Functors:

    3 constant functors

    2 with the empty set on object 1, and the remainder either on object 2 or 3

    1 with the empty set on object 2 and the remainder mapped to object 3

    Comment Source:since there's always a function from the empty set to any set but not the other way around, we are down to a total of 6 Functors: 3 constant functors 2 with the empty set on object 1, and the remainder either on object 2 or 3 1 with the empty set on object 2 and the remainder mapped to object 3
  • 4.

    For a bit I mistook the category of "Set" for the category of "Set and Injections". Which begs the question, how many functors are there "Set and Injections" -> 3.

    In "Set and Injections" you can have the singleton sets serve as a sort of dual to the empty set and create four more functors in addition to the ones proposed by @FabricioOlivetti :

    2 with the singleton sets on object 3, and the remainder either on object 1 or 2

    1 with the singleton sets on object 2, and the remainder mapped to object 1

    1 with the empty set on object 1, the singleton sets on object 3, and the remainder mapped to object 2

    Comment Source:For a bit I mistook the category of "Set" for the category of "Set and Injections". Which begs the question, how many functors are there "Set and Injections" -> 3. In "Set and Injections" you can have the singleton sets serve as a sort of dual to the empty set and create four more functors in addition to the ones proposed by @FabricioOlivetti : 2 with the singleton sets on object 3, and the remainder either on object 1 or 2 1 with the singleton sets on object 2, and the remainder mapped to object 1 1 with the empty set on object 1, the singleton sets on object 3, and the remainder mapped to object 2
Sign In or Register to comment.