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Suppose $C$ is a complete and cocomplete cartesian-closed category and $G$ is a group object in $C$.

Define $G$-$C$ as the category having objects the objects $X$ of $C$ together with a $G$-left-action $\sigma:G\times X\to X$ and equivariant $C$-morphisms as morphisms.

I wonder what the product and the coproduct in the category $G$-$C$ are. My guess is that the action of the product of two objects $(X,\sigma)$, $(X',\sigma')$ of $G$-$C$ is given by $$ G\times (X\times X')\xrightarrow{\Delta\times id}(G\times G)\times(X\times X')\cong(G\times X)\times (G\times X')\xrightarrow{\sigma\times\sigma'}X\times X' $$ and that the action of the coproduct is given by $$ G\times (X\coprod X') \cong (G\times X)\coprod(G\times X') \xrightarrow{\sigma\coprod\sigma'} X\coprod X'. $$ For the isomorphism $G\times(X\coprod X')\cong (G\times X)\coprod(G\times X')$ I need a cartesian-closed $C$. My question is: Does this really define the product and the coproduct on $G$-$C$?

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Short answer: Yes! Just check it. By the way, it also works when $G$ is a monoid object. –  Martin Brandenburg Sep 4 '13 at 22:52

1 Answer 1

up vote 3 down vote accepted
  1. Let $\mathcal{C}$ be a category with finite products and let $G$ be an internal group in $\mathcal{C}$. Then the category $[G, \mathcal{C}]$ of $G$-objects in $\mathcal{C}$ is monadic over $\mathcal{C}$, i.e. its forgetful functor to $\mathcal{C}$ has a left adjoint and $[G, \mathcal{C}]$ is isomorphic as a category over $\mathcal{C}$ to the category of algebras for that monad. In particular, $[G, \mathcal{C}]$ has whatever limits $\mathcal{C}$ has, and they are computed as in $\mathcal{C}$.
  2. If $\mathcal{C}$ is cartesian closed, then the forgetful functor $[G, \mathcal{C}] \to \mathcal{C}$ has a right adjoint (namely the functor $X \mapsto X^G$ with the induced $G$-action) and is moreover comonadic. Thus in this case $[G, \mathcal{C}]$ has whatever colimits $\mathcal{C}$ has, and they are computed as in $\mathcal{C}$.

Thus in a cartesian closed category, $[G, \mathcal{C}]$ behaves a lot like what one would expect from the $\mathcal{C} = \mathbf{Set}$ case. In fact, $[G, \mathcal{C}]$ even inherits the property of cartesian closedness. (If $X$ and $Y$ are $G$-objects, the exponential object $Y^X$ has a left $G$-action given by $(g \cdot f) (x) = g \cdot f (g^{-1} \cdot x)$, and one may verify that this makes $Y^X$ an exponential object in $[G, \mathcal{C}]$.)

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Dear @Zhen Lin, thanks for the answer. The first two points tell me as far as I understand only that the underlying $C$-object of the product in $G$-$C$ is the product in $C$ and analogously for the coproduct. The question however was about the actions. –  Ronald Bernard Sep 4 '13 at 22:57
The actions are constructed as you say. –  Zhen Lin Sep 4 '13 at 23:04

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