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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?

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  • $\begingroup$ 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
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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.

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