To explain the question, I'd like to start by considering monoid actions in $\mathbf{Sets}$. In this case, a monoid action is simply a functor $S$ from a monoid $M$ to $\mathbf{Sets}$. Then, one can form a category of monoid actions, wherein objects are functors $S \colon M \to \mathbf{Sets}$, and wherein morphisms between $S \colon M \to \mathbf{Sets}$ and $S' \colon M' \to \mathbf{Sets}$ (note that we allow the monoid to be different in $S$ and $S'$; in other words, this is not the category $\mathbf{Sets}^{M}$) are pairs $(L,\lambda)$ where $L$ is an monoid homomorphism $L \colon M \to M'$ and $\lambda$ is a natural transformation $\lambda \colon S \to S'L$. Then, we can define the automorphism group of a functor $S \colon M \to \mathbf{Sets}$ based on the set of pairs $(L,\lambda)$ where $L$ is an isomorphism and $\lambda$ is an equivalence.

If one now considers $M$ to be a monoid of binary relations on a finite set, we get a functor $S \colon M \to \mathbf{Rel}$, where $\mathbf{Rel}$ is the category of finite sets and binary relations between them. The question is: how should the automorphisms of $S$ be defined ?

The simple choice would be to consider pairs $(L,\lambda)$ as above, where $\lambda$ is a natural transformation, i.e. for any relation $\mathcal{R}$ on a set $X$, $x \mathcal{R} y \iff \lambda(x) L(\mathcal{R}) \lambda(y)$.

On the other hand, $\mathbf{Rel}$ is a 2-category, so one could also consider $\lambda$ to be a lax natural transformation, allowing inclusions of relations.

What would be the proper way to define the automorphism group of $S$ ?

  • 2
    $\begingroup$ Depends on your application. Either might end up being appropriate. $\endgroup$ – Qiaochu Yuan May 10 '16 at 20:30
  • $\begingroup$ @Qiaochu Yuan: I am realizing now that if I ask $L$ to be an isomorphism, and $\lambda$ an equivalence (hence a bijection), then whatever 2-morphism which should be present in the lax natural transformation should be invertible and thus, given the nature of 2-morphisms in $\mathbf{Rel}$, should be the identity. Am I correct ? $\endgroup$ – OliverX1 May 16 '16 at 19:23

Your Answer

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

Browse other questions tagged or ask your own question.