An old <a href="http://mathoverflow.net/questions/1365/is-there-a-universal-group-object-answered-yes/1477#1477">MO answer</a> 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?