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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.

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Presumably, F fixes the terminal object 1, so Hom(1, G) = Hom(F(1), F(G)) = Hom(1, F(g)), right? –  Ryan Reich May 30 '11 at 17:36
    
Oh, right. That makes sense. –  Qiaochu Yuan May 30 '11 at 17:39
    
Note that the example with (not necessarily commutative) rings is just like your vector-space example, in that the antimorphisms are the contravariant functors: specifically, viewing a ring as a one-object, abelian-group enriched category, the desired anti-morphisms between rings are the contravariant enriched functors (as the notion of "opposite ring" is just a special case of the general notion of "opposite category"). –  Sridhar Ramesh May 30 '11 at 18:08
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I also want to note that inversion is only seemingly a structure even when given as a morphism. If you interpret the structure of a monoid object as being a factorization of the functor of points through the category of monoids, then the property of being a group object is the property of that factorization taking values in the full subcategory of groups. This gives a natural inversion on the functor of points and, thus, on the object itself. On the category side, the inversion map (if it exists) is unique, so its existence is merely a property. –  Ryan Reich May 30 '11 at 18:22
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My first reaction to this is to say: "You silly twisted boy, you". –  Loop Space May 30 '11 at 21:30
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2 Answers

up vote 4 down vote accepted

$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\id}{\mathrm{id}}$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.

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Oh, I see. This is quite nice. We need $S$ to preserve products, though, right? –  Qiaochu Yuan May 30 '11 at 19:30
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By the way, when we discussed this (actually at tea) years ago, I think we decided an "anti-morphism" should obviously be called a "heteromorphism." –  Ben Webster May 30 '11 at 19:56
    
@Qiaochu: Yes. Basically, I have in mind that $\cat{C}$ is a "concrete in $\cat{D}$ category", like Poisson manifolds in sets or some such. @Ben: Of course! It's perfect. I shall edit immediately. –  Ryan Reich May 30 '11 at 20:22
    
There's also an obvious concept of a "weak $(\cat{C}, S)$-group object", for lack of a better term, in which we give a monoid in $\cat{C}$ which is a group in $\cat{D}$ but with its inverse not necessarily lifting at all. I have no idea what that entails. –  Ryan Reich May 30 '11 at 20:33
    
The conceptual difficulty that motivated the first question has passed, so I don't need an example anymore. The situation is just that sometimes group objects in the usual sense form too small a collection of examples to include the ones of interest, so one weakens the definition as above. For example apparently all group objects in the usual sense in the category of Poisson manifolds have trivial Poisson bracket, and this is a shame, so one forgets the bracket. –  Qiaochu Yuan May 31 '11 at 15:03
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Here's a repair of the definition suggested in the OP. Namely, we need the additional assumption that $\text{Hom}(1, -) : C \to \text{Set}$ is faithful (so concretizes $C$). Then with the condition described in the OP, we get a map of sets $^{-1} : \text{Hom}(1, G) \to \text{Hom}(1, G)$, and now it really makes sense to ask whether this lifts to a morphism in $C$; faithfulness ensures that if such a morphism exists, it is unique, and therefore really a property of $G$.

So it may happen that $^{-1}$ exists but is not a morphism in $C$ (and that this is the sense in which $G$ is a group object), but also that $\text{Hom}(1, -)$ factors through some other category $D$ in which it is a morphism. For example, if $C$ is the category of Poisson manifolds, then apparently $D$ is the category of smooth manifolds. This provides some motivation for the definition in Ryan Reich's answer, and actually it suggests an even weaker definition (edit: which I see Ryan has already suggested!):

Let $C, D$ be categories with finite products and $S : C \to D$ a faithful, product-preserving functor. A (let's say) $D$-virtual group object is a monoid object $G \in C$ together with a morphism $i : S(G) \to S(G)$ such that $S(G)$ (with the induced monoidal structure) is a group object in $D$ with inverse $i$.

I don't know whether this implies that there needs to be a good notion of heteromorphism associated to $i$ (unless by "heteromorphism" you mean nothing more and nothing less than a morphism in $D$).

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Actually, it sounds like you are really motivating my last comment to my answer. So you do think it is meaningful to consider an inversion map which simply has no analogue in $\mathbf{C}$? –  Ryan Reich May 31 '11 at 15:49
    
Ha, comment conflict. What you just edited in is also what I was suggesting. –  Ryan Reich May 31 '11 at 15:50
    
Oh, ha again. I thought I said something different, but that really is the last comment in your answer. Okay, before I say anything else, can someone tell me whether there's any sense in which $i^2$ must be a morphism? –  Qiaochu Yuan May 31 '11 at 16:03
    
I think the "virtual group object" setting is the only appropriate one in which to interpret this question, because it emphasizes that the inverse only exists in $\mathbf{D}$. In that sense, $i^2$ is the identity in $\mathbf{D}$ like for any group object, and thus lifts to $\mathbf{C}$ (as the identity, since $S$ is faithful). If you happen to be working with some kind of inversion $F$, and if $F^2 = I$, then you could also say $s(i)i = \mathrm{id}$, but I wonder if it is possible to have, say, an $F$ of order 3, so that we only have "$i^3 = \mathrm{id}$". An example would be nice here. –  Ryan Reich May 31 '11 at 16:16
    
I meant $F(i) i$. –  Ryan Reich May 31 '11 at 16:17
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