What is a cogroup and what are coactions?

Question

What is a cogroup and what are coactions?

A very nice way to think about a group action on an object $X$ is as a group homomorphism from $G$ to $\operatorname{Aut}(X)$. Is there something similar for cogroups?

Answer 1

A co-group is the dual of a group, and a co-group action is the dual of a group action. Precisely:

A group $G$ is a pointed set (more generally, an object in a cartesian category) equipped with maps $\mu: G\times G \to G$ (the multiplication), $\eta: \ast \to G$ (the unit), and $i: G\to G$ (the inverse), satisfying the usual group axioms. These axioms can be encoded by asking certain diagrams to commute. Those diagrams are drawn on the nLab. For a cartesian category $\mathcal{C}$, the object denoted by $\ast$ is the initial object. Think of the map $\eta$ as picking one special element of $G$ that we will denote by $1$.

For a cogroup $C$, just reverse the directions of all the arrows, replace product by coproduct (in $Sets$, or more generally any cocartesian monoidal category), and keep the same diagrams. So now there's a comultiplication $\Delta: G\to G\sqcup G$, a counit $\nu: G\to \{0\}$, and a co-inverse $i: G\to G$. When you reverse the associativity diagram we call it co-associativity. There's also a co-unit axiom (reversing the unit axiom that says $g\cdot1 = g = 1\cdot g$) and a co-inverse axiom (reversing the inverse axiom that says $g\cdot g^{-1} = 1 = g^{-1}\cdot g$). In a general cocartesian monoidal category, the object denoted $\{0\}$ above is the initial object, i.e., the coproduct of zero objects. Technically, we don't need all coproducts to exist. We need the coproduct of zero objects, the coproduct of two objects (for the $G\sqcup G$ above), and $G\sqcup G\sqcup G$ (for the coassociativity axiom).

A group-action is an object $X$ together with an action map $m: G\times X \to X$, satisfying the usual axioms, e.g., to say $g\cdot(h\cdot x) = (gh) \cdot x$ can be encoded by a certain diagram commuting. By hom-tensor duality, an element $m$ in $Hom(G\times X,X)$ is the same as an element in $Hom(G,Hom(X,X))$, i.e., a homomorphism from $G$ to $Aut(X)$.

For a co-group action, simply reverse all the arrows and replace product by coproduct. So now you have a co-action map $c: X\to G \sqcup X$ satisfying the reversed diagram from above. This $c$ is an element of $Hom(X,G\sqcup X)$. Unfortunately, the left adjoint is on the wrong side. It will not be true in general that a co-group object is a map from $Aut(X)$ to $G$.

Answer 2

I think to complete David White's answer it is also interesting to mention some stuff :

A group in a category $\mathcal{C}$ with products is as David said an object with some data involving only finite products.

One can say cogroup object in a category $\mathcal{C}$ with finite coproducts is a group object in $\mathcal{C}^{\operatorname{op}}$ and you get David's description with coproducts.

(With these definitions, you can see that if $G$ is a group, then for any $X$, $\operatorname{Hom}(X,G)$ is a group in $Set$ and if $G$ is a cogroup, then for any $X$, $\operatorname{Hom}(G,X)$ is a group in $Set$, this will be relevant for an example later)

But I think a better setting would be to fix a unital monoidal category $(\mathcal{C},\otimes,\mathbf{1})$ where you have diagonals (that is, for any $X$, a map $X \rightarrow X \otimes X$ that is natural and compatible with the monoidal category structure, see here for the axioms. This assumption is important for the inverse axiom because you want to talk about elements of the for $x\otimes x$ for $x\in X$ (and then talk about $i(x)\otimes x$). In a general monoidal category the existence of such diagonal map is not automatic) and to say that a group object is the data of an object $G$ with maps $G \otimes G \rightarrow G$, $\mathbf{1} \rightarrow G$, and $G\rightarrow G$ called respectively the mulitplication, the unit, the inverse and such that they make some diagram commute (corresponding to the unit, the associativity, and the inverse axioms) as in the nlab page. For instance taking group objects in $(Set,\times,*)$ you recover the classical notion of groups.

A group action of a group object $G$ on $X$ would be the data of a map $G\otimes X \rightarrow X$ satisfying some diagrams that commutes provided on the nlab page. Again for $Set$ you recover the classical notion of actions.

Once this settled up, you can say that a cogroup object in $(\mathcal{C},\otimes,\mathbf{1})$ is a group object in $(\mathcal{C}^{\operatorname{op}},\otimes,\mathbf{1})$ with the same monoidal structure, but here you need to assume you have codiagonals (maps $X \otimes X \rightarrow X$), and a cogroup action is a group action in $(\mathcal{C}^{\operatorname{op}},\otimes,\mathbf{1})$. This does not fit in the first approach of cogroup I have given but it is more general because in that setting you can talk about groups/cogroups for monoidal structures other that the cartesians one. You can write what it is explicitely by taking the axioms of a group object and taking the opposite of all arrows.

Before giving some example I have a little warning to make. In a closed monoidal category, the tensor product $\otimes$ has a right adjoint $\mathcal{Hom}$ called the internal Hom. For instance in $Set$ the Hom set in the internal Hom. I think you can prove (maybe with some extra assumptions) that $\mathcal{Hom}(X,X)$ is always a monoid object (the same as a group object but you don't ask for an inverse map). In that case, for group actions, you can dualize the data $G\otimes X \rightarrow X$ into a data $G \rightarrow \mathcal{Hom}(X,X)$. Asking for $G \otimes X \rightarrow X$ to satisfy action axioms is exactly the same as asking $G \rightarrow \mathcal{Hom}(X,X)$ to be a monoid homomorphism. And you recover the definition of group action you were referring to.

But for cogroups, in general you can't say $\mathcal{Hom}$ is left adjoint to $\otimes$ so you can't really talk about cogroups that way (or at least I don't know any good examples of object being "cogroups" using the $\mathcal{Hom}$ characterization or with a left adjoint to $\otimes$ characterization).

Here are some examples of cogroups :

• Hopf algebras are special case of cogroups in the category of unital rings.

• In $(hTop_*,\vee)$ (Pointed topological spaces with maps between homotopy classes of pointed continuous maps, and $\vee$ the smash product), the spheres $S^n$ with a fixed points are cogroups. The counit $S^n \rightarrow *$ being the terminal map, the coinverse is the symmetry with the equator $S^{n-1} \subset S^n$ and the comulitplication is $S^n \rightarrow S^n \vee S^n$ is the pinching operator sending $S^n$ to $S^n/S^{n-1} \simeq S^n \vee S^n$. You can moreover see that for $(X,x)$ a pointed topological space, $\operatorname{Hom}_{hTop_*}(S^n,(X,x)) = [S^n,(X,x)]$ is a group. In fact, that's a way to define $\pi_n(X,x)$.

• For coactions maybe you can first see that monoid objects in $\mathbb{Z}$-$\textrm{Mod}$ with the tensor product, are rings. And (left) actions of rings are left modules over rings. If $G$ is a finite group, I think the function ring $\mathcal{C}(G,\mathbb{R})$ would be a (commutative) Hopf algebra, and in particular a cogroup in $\mathbb{R}$-$\textrm{Mod}$. And modules that have a coaction on this are exactly representations of $G$.