Automorphisms of $\bf{Set}$ are naturally isomorphic to the identity

Question

I am trying to prove that if $F$ is an automorphism of the category $\bf{Set}$, then there is a unique natural isomorphism from the identity functor to $F$.

By Yoneda Lemma, there is a bijection between natural transformations $\mathcal{C}(A,-)\longrightarrow F$ in $[\bf{Set},\bf{Set}]$ and elements of $FA$, where $\mathcal{C}(A,-)$ is the functor sending $B$ to $\mathcal{C}(A,B)$ (set of morphisms from $A$ to $B$).

I believe that by applying Yoneda Lemma taking $A=\varnothing$, the result can be proved. However, I am quite confused because I do not know how to manage functions having domain $\varnothing$.

Can anyone help me, please?

Answer 1

First of all, the result is false in general: consider the functor sending every set to $\emptyset$ and every morphism to $id_\emptyset$. So presumably you want a further condition on $F$.

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You want two things here:

• You want $\mathcal{C}(A, -)$ to be the identity (since you're looking for a natural transformation from the identity to $F$).

• You want $FA$ to have a single element (since you want there to be exactly one natural transformation).

Taking $A=\emptyset$ fails both bulletpoints:

• For any set $X$, there is a unique map $\emptyset\rightarrow X$, namely the empty map. (In jargon: $\emptyset$ is the initial object in Sets.) So $\mathcal{C}(\emptyset, -)$ sends every set to a one-element set. That's certainly not the identity functor!

• We may well have $F\emptyset=\emptyset$. In this case, $F\emptyset$ does not have exactly one element.

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So $A=\emptyset$ won't work. What will?

Well, as noted above, we need some further conditions on $F$ to make our statement true, but we can still figure out what $A$ should be. Both the following questions suggest a guess for $A$ (up to isomorphism, anyways):

• We want $\mathcal{C}(A, B)$ to "basically" be $B$. That is, maps from $A$ to $B$ should be "basically" elements of $B$. $\emptyset$ doesn't work here, since there are too few maps from $\emptyset$ to $B$ in general. Something like $\{0, 1\}$ won't work, since in general there are too many maps from $\emptyset$ to $B$ (e.g. take $B=\{0, 1\}$ - $Hom(\{0, 1\}, \{0,1\})$ has four elements, not two). What sort of set should $A$ be so that maps from $A$ to $B$ correspond to individual elements of $A$?

The "elements-as-morphisms-from-an-appropriate-object" is an important idea on its own, by the way.

• We want $FA$ to have exactly one element. Based on that, how many elements do you think $A$ should have?