Hopf monoids in different categories

Question

Hopf algebras are precisely Hopf monoids in the category of vector spaces.
What are Hopf monoids in other common (to be interpreted by the reader) monoidal categories?

In particular, are Hopf monoids in the cartesian monoidal category $(Set, \times, *)$ precisely groups?

One could imagine the antipode to correspond to taking the inverse, the coproduct to the diagonal map, and the product to the multiplication in the group. Further, the counit could be seen as an augmentation. And so on ...

Why am I asking? I want to enhance my understanding of Hopf algebras.
Above all, I want to compare them to algebraic structures I have encountered already.

Answer 1

As you probably already know, a monoid object $M$ in the category of sets with Cartesian product is precisely a monoid in the usual sense: the unit map $u: * \to M$ takes the single-element set to the unit element of the monoid, while the multiplication $m \colon M \times M \to M$ gives the monoid multiplication.

What is a comonoid object $C$ in this category? There is only a single possibility for the counit map $\epsilon \colon C \to *$, so we need to only work out the comultiplication $\Delta \colon C \to C \times C.$ Since $\times$ is the categorical product, $\Delta$ is equivalent to two maps $l: C \to C$ and $r: C \to C$ such that $\Delta(c) = (l(c), r(c))$ for all $c \in C$.

The left counit axiom says that $c \mapsto (*, c)$ should be equivalent to $c \mapsto (l(c), r(c)) \mapsto (*, r(c))$. Hence we find that $r: C \to C$ must be the identity function. Similarly for $l$, and so $\Delta \colon C \to C \times C$ must simply be the map $\Delta(c) = (c, c)$. It is easy to see that $\Delta$ is coassociative.

Checking the bimonoid axioms, every monoid object $(M, u, m)$ automatically becomes a bimonoid object $(M, u, m, \epsilon, \Delta)$ with $\epsilon$ and $\Delta$ as above. So all that is left is to figure out what antipode map $S \colon M \to M$ needs to satisfy. One of the Hopf axioms is that the composition $$ x \mapsto (x, x) \mapsto (x, S(x)) \mapsto m(x, S(x))$$ is equal to the composition $$ x \mapsto * \mapsto 1$$ where $1$ is the unit in the monoid. So $S(x)$ must be a right inverse for $x$. The other Hopf axiom gives that $S(x)$ is a left inverse for $x$.

So indeed, a monoid object in the category of sets is always a bimonoid object, and this can be equipped with an antipode if and only if every element of the monoid is invertible.

Answer 2

I just want to add to Joppy's answer by pointing out that once we know Hopf monoids are group objects in $\newcommand\Set{\mathbf{Set}}\Set$, then we get for free that Hopf monoid objects in any cartesian monoidal category $(C,\times,*)$ are precisely the group objects.

The argument goes by the Yoneda lemma.

Let $(M,u,m,\epsilon,\Delta,S)$ be a Hopf algebra object in $C$. You can check for any $x\in C$, applying the functor $\newcommand\Hom{\operatorname{Hom}}\Hom(x,-)$ to this data gives a Hopf algebra object in $\mathbf{Set}$. This is essentially because $\Hom(x,-)$ preserves limits, and in particular products, i.e., $\Hom(x,M\times M)\simeq \Hom(x,M)\times \Hom(x,M)$. Moreover this Hopf algebra structure is functorial in $x$.

But we already know that Hopf monoids in $(\Set,\times, *)$ are the same as groups. Therefore, the functor $\Hom(-,M)$ has a natural group structure, which by Yoneda reflects to a natural group structure on $M$.

The same argument goes in reverse as well, a group object gives a functor to groups in $\Set$, which are also naturally Hopf monoids, and thus we get a Hopf monoid object structure on the original object.