Let $S$ be a semigroup and $X$ a set. A left $S$-action structure on $X$ is a semigroup arrow $S\to \mathsf{Set}(X,X)$.

Let $M$ be an abelian group and $R$ a ring. A left $R$-module structure on $M$ is a ring arrow $R\to\mathsf{Ab}(M,M)$.

In both cases we are defining actions of semigroup/monoid objects in a category $\mathsf C$ within the category of internal monoids $\mathsf{Mon}(\mathsf C)$.

Usually, equivariance is defined in the category $\mathsf {C}$ itself using diagrams in $\mathsf {C}$ involving the monoid structure arrows of the acting object. Can we define equivariance in $\mathsf{Mon}(\mathsf {C})$?

For instance, can we define $R$-linear arrows of $R$-modules in the category of rings? Perhaps in the commutative case?

  • $\begingroup$ How are you planning to make an arbitrary ring into an R-module? $\endgroup$ – Pedro Tamaroff Oct 25 '18 at 15:08
  • $\begingroup$ @PedroTamaroff I'm not sure I understand your question. I think one can apply the forgetful functor $U$ from rings (monoid objects in abelian groups) to abelian groups and use the structure maps to give a ring arrow $R\to \mathsf{Ab}(UR,UR)$. $\endgroup$ – Arrow Oct 25 '18 at 15:12

If the ambient category is monoidal closed, the usual equivariance conditions may be expressed with internal homs. This expression does not live in $\mathsf{Mon}(\mathsf C)$, but I thought it's still worth recording.

Let $M$ be an internal monoid and consider actions $M\to \underline{\mathsf{C}}(A,A)$ and $M\to \underline{\mathsf{C}}(B,B)$. Observe any arrow $f:A\to B$ in $\mathsf C$ induces morphisms by pre-composition and post-composition. This arrow is equivariant precisely if the following diagram commutes.

$$\require{AMScd} \begin{CD} M @>>> \underline{\mathsf{C}}(A,A)\\ @VVV @VV{f_\ast}V\\ \underline{\mathsf{C}}(B,B) @>>{f^\ast}> \underline{\mathsf{C}}(A,B) \end{CD}$$

However, this still takes place in the category $\mathsf C$ and not $\mathsf{Mon}(\mathsf C)$ since the bottom right corner has no monoid structure.


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.