An old MO answer by Noah Snyder makes a claim I don't completely understand, but mostly because I don't know any examples. The answer claims that in some examples of (things that one would want to call) group objects $G$ in some category $C$ with finite products, inversion is not a morphism, but an "anti-morphism" (if that notion makes sense in $C$).
- Why should this be the case? (Preferably illustrated with a simple example)
- What does "anti-morphism" even mean in general?
Here's a stab at the second question: if $C$ is equipped with a functor $F : C \to C$ with $F^2 \cong \text{id}_C$, then perhaps an anti-morphism $G \to G$ is a morphism $G \to F(G)$ (or equivalently a morphism $F(G) \to G$). In the category of Poisson manifolds $F$ appears to correspond to negating the Poisson bracket, whereas in the category of noncommutative rings $F$ appears to correspond to taking the opposite ring.
But $F$ ought to be special in some way since, as Noah says, inversion is a property, not a structure. I guess what he means by this is that the "correct" definition of a group object is
A monoid object $G$ such that for every point $g : 1 \to G$ there is a unique point $g^{-1} : 1 \to G$ such that the composite of $g \times g^{-1} : 1 \to G \times G$ with multiplication $m : G \times G \to G$ gives the identity $e : 1 \to G$, and similarly in the other order.
But this seems possibly too weak of a definition if there aren't many morphisms $1 \to G$. In any case, the map sending $g$ to $g^{-1}$ is a map of sets $\text{Hom}(1, G) \to \text{Hom}(1, G)$, and I guess this ought to be lifted to an anti-morphism $G \to F(G)$ in some way, which I suppose means we need a natural identification $\text{Hom}(1, G) \cong \text{Hom}(1, F(G))$. And this seems like an oddly specific thing to demand of $F$ unless $F$ canonically arises somehow from some other procedure (that is, is itself a property of, and not a structure on, $C$). Is that the case in the above examples? See Ryan Reich's comment below.
Edit: Is the following an example? Let $\text{Vect}$ be the category of finite-dimensional vector spaces over a field. We'd like to be able to say that $\text{Vect}$ is some kind of "really weak group object" in $\text{Cat}$ in the sense that it's got a multiplication $\otimes$ given by tensor product and a "weak inverse" given by taking the dual space, but taking dual spaces is contravariant. So I guess contravariant functors are the "anti-morphisms" in $\text{Cat}$, which means that $F$ is taking the opposite category.
Best Answer
$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\id}{\mathrm{id}}$I like your definition of an antimorphism (which, following Ben Webster's suggestion, I will call a "heteromorphism") and I'll raise you one: if $\cat{C}$ comes with an autoequivalence $F$, an $F$-heteromorphism is by definition a morphism $X \to F(X)$. Why work in this generality? Here's why.
I see one big issue with defining a group with heteromorphism inverse, namely, how one is to state the inversion property: $$G \xrightarrow{\Delta} G \times G \xrightarrow{i \times \id} G \times G \xrightarrow{m} G$$ if in fact we have $i \colon G \to F(G)$; how can we get both of the latter factors the same so that $m$ may be applied? My answer is, philosophically: since a heteromorphism is a morphism after allowing the loss of some structure, we must check this diagram also after forgetting that structure. This will turn $F$ into the identity.
Here's what I mean. Let $S \colon \cat{C} \to \cat{D}$ be some "structure-forgetting" faithful functor that preserves products, and suppose $F \colon \cat{C} \to \cat{C}$ acts fiberwise for $S$, in that $SF = S$ (I suppose more generally we could also specify $\phi \colon SF \cong S$). For example, $S$ could be "forgetting the Poisson structure" or "forgetting the multiplication in a noncommutative ring" or "forget the directions of arrows in categories". Correspondingly, $F$ would be "take the negative Poisson structure" or "take the opposite ring" or "take the opposite category". We will define all the morphisms for a group object in $\cat{C}$, but check their properties in $\cat{D}$ (since $S$ is faithful, this won't result in any errors).
So, say that a $(\cat{C},F)$-group object (any suggestions for a better name for this?) is an object $G \in \cat{C}$ together with morphisms $$m \colon G \times G \to G, \qquad i \colon G \to F(G), \qquad u \colon 1 \to G$$ constituting a group object structure on $S(G)$. In particular, the above diagram reads $$S(G) \xrightarrow{\Delta} S(G) \times S(G) \xrightarrow{S(i) \times \id} SF(G) \times S(G) \xrightarrow{m} S(G),$$ where we have $SF = S$ by definition.
Note that the inverse $i$, if it exists, is unique, since $S(i)$ is unique in $\cat{D}$ and $S$ is faithful. So it is, as for regular group objects, not a structure but a property. Note also that I omit the notation $S$ from "$(\cat{C},F)$-group object" because it is enough that some $S$ exist and that $F$ act on its fibers, but it doesn't matter which one we use.