Pretty much anyone who does algebra is familiar with group objects in categories, but what about cogroup objects? Most of what I've been able to find about them is that they "arise naturally in algebraic topology" (from wikipedia) and that, somehow, the nsphere is one (nLab's meager entry). Is there a reference for more on the stuff? Specifically wondering if the spaces of pointed maps from a topological space X to a pointed sphere are cogroups, and if anything is known about these "cohomotopy groups."

1$\begingroup$ Of course, anyone who's ever used the diagonal map of anything has thought about cogroups... $\endgroup$ – Ben Webster♦ Nov 2 '09 at 4:04

1$\begingroup$ Full cogroups? Or possibly something weaker like a comonoid, possibly? $\endgroup$ – Mikael VejdemoJohansson Nov 2 '09 at 5:31

10$\begingroup$ did you know that there is a rule that says that anyone who calls an nLab entry "meager" is required to expand it? $\endgroup$ – Urs Schreiber Nov 3 '09 at 13:18

5$\begingroup$ as if by magic, a nonmeager discussion of cogroups has now appeared here: ncatlab.org/nlab/show/cogroup+object $\endgroup$ – Urs Schreiber Nov 3 '09 at 14:34

$\begingroup$ Oooh, very nice. I was going to add one in a few days after I was sure the discussion here settled down, but glad someone else did it first. $\endgroup$ – Charles Siegel Nov 3 '09 at 17:42
Spheres are (homotopy) cogroups for the same reason that homotopy groups are groups. The comultiplication S^n \to S^n \vee S^n is the map that collapses the equator, the same map that is used to define composition in homotopy groups. Note that this only satisfies the cogroup axioms up to homotopy, just as composition in the homotopy group is only a group operation because you are taking homotopy classes.
However, Maps(X,S^n) does not inherit a natural cogroup structure from S^n, because Maps(X,S^n \vee S^n) does not naturally map to Maps(X,S^n) \vee Maps(X,S^n). However, there is something called (stable) cohomotopy groups: these are stable homotopy classes of maps from X to S^n. This is a cohomology theory, the cohomology theory associated to the sphere spectrum. This cohomology theory is very hard to compute though; its value on a point is the stable homotopy groups of spheres!
The way you can get a cogroup from a sphere and an arbitrary space is by taking smash products, rather than mapping spaces: for any X, X \wedge S^n (which is the nfold suspension of X) is a homotopy cogroup. You can see this by using the same "collapse the equator" map on the suspension, or you can see it more categorically from the fact that we have a map X \wedge S^n \to X \wedge (S^n \vee S^n) = (X \wedge S^n) \vee (X \wedge S^n). More generally, if C is a cogroup, so is X \wedge C for any X (and Maps(C,X) is a group), and if G is a group, so is Maps(X,G) for any X.

6$\begingroup$ "\wedge" is a somewhat confusing name in this context :) $\endgroup$ – Reid Barton Nov 2 '09 at 4:30

$\begingroup$ No, it isn't. \wedge is the LaTeX symbol that is the reflection from \vee: \vee = \/, and \wedge = /\. $\endgroup$ – Mikael VejdemoJohansson Nov 2 '09 at 5:34

9$\begingroup$ \vee describes the wedge operation, while \wedge describes smash product. $\endgroup$ – S. Carnahan♦ Nov 2 '09 at 5:56

1$\begingroup$ @S.Carnahan only to homotopy theorists. to everyone else it's still just a wedge $\endgroup$ – Jonathan Beardsley Jun 28 '13 at 1:04
You may be thinking maps into a cogroup object should form a cogroup the way maps into a group object form a group. This isn't quite true; the "correct" analog is that maps out of a cogroup object form a group.
More precisely, if G is an object in a category and D is its group object diagram, then applying the functor Hom(X,) to D yields a group object diagram making Hom(X,G) into a group. Essential here is the fact that Hom(X,) preserves products (in fact all limits); it does not generally preserve coproducts, which is why in general one would not expect Hom(X,S) to inherit a cogroup structure from a cogroup structure on S. Hopefully this addresses your ponderings about the possibility of "cohomotopy groups".
The dual statement is true for cogroups: if S is a cogroup, applying Hom(,X) to its cogroup diagram yields a diagram defining a group structure on Hom(S,X). Here, dually, it's essential that Hom(,X) turns coproducts into products (in fact all colimits into corresponding limits); it does not generally turn products in coproducts.
This makes precise the relationship of the cogroup structure on the spheres to homotopy groups: if you work through the example of S^{n} and see how its cogroup structure makes Hom(S^{n,},X) into a group after applying Hom(,X), you'll get exactly the nth homotopy group of X :)
Eric already explained exactly how the suspension of a space and in particular S^{1} is a cogroup object in the homotopy category of pointed spaces. I'll just point out that we know it must be one, by Yoneda, because the functor it corepresents [S^{1}, –] = π_{1} is groupvalued.
This is less of an answer and more of a bit of bickering about terminology. I think that the word "group object" rightly means the following:
A group object G in a Cartesian category is a map G x G → G, satisfying an associativity axiom, such that there exists a map 1 → G satisfying the unit axiom (it is then necessarily unique) and a map G → G satisfying the inversion axiom (it is then necessarily unique).
If you want, you can consider the unit and inversion maps as part of the data. What's important in the word "group object" is that it is in a Cartesian category, i.e. a category with products. The object 1 is any choice of terminal object for the category.
From this perspective, the most obvious notion of "cogroup object" is:
A cogroup object G in a Cartesian category is a map G → G x G, satisfying a coassociativity axiom, such that there exists a map G → 1 satisfying the counit axiom (it is then necessarily unique) and a map G → G satisfying the inversion axiom (it is then necessarily unique).
But this is a rather problemmatic definition. First of all, by definition there is a unique map G → 1 for any object G. Second, there is a unique coassiciative map G → G x G; it is given by the diagonal map, which is the element of Hom(G,GxG) corresponding to (id,id) ∈ Hom(G,G) x Hom(G,G). (Remember that the definition of Cartesian product is that for any H there is a natural isomorphism Hom(H,GxG) = Hom(H,G) x Hom(H,G).) So every object of a Cartesian category is a "counital cosemigroup" in a unique way. Finally, it's rather hard to write down exactly what the inversion axiom should say.
Well, so the problem is that we're trying to use the Cartesian structure. So Wikipedia and elsewhere adopt the only nontrivial definition, which is to say that for cogroups you should use the coproduct, not the product, so you should ask for a map G → G + G. I think this is the question you're asking about. I guess I would word it "are the any natural categories C with interesting group objects in C^{op}?", since this definition of "cogroup object" really is "group object in the opposite category".
Finally, I'll mention that certain related constructions do show up all the time. There are good words "algebra" and "coalgebra", which refer to objects in categories with designated monoidal structures (e.g. tensor product in VECT). What's cool is that to make sense of the inverse map's axiom when you aren't using either the product (for groups) or the coproduct (for cogroups), you actually need all the data of a "bialgebra". So the correct notions of "group" and "cogroup" coincide when you give up on categorical (co)products, and become the word "Hopf algebra".

2$\begingroup$ Yes: "co""something" corresponds to "something" in the opposite category. This usage is very consistent: coproducts, colimits, cogroups, comonoids, coalgebras... so I'd say the coproduct definition of cogroup really is the obvious one. $\endgroup$ – Andrew Critch Nov 2 '09 at 7:41

$\begingroup$ So correct me if Im wrong.. for cogroup objects we use comultiplication morphism which brings the object G to the coproduct G U G .. but what bothers me is the identity morphism.. in cogroups if it is really a dual notions.. then shouldnt 1 be the initial object of the category.. which gives me the implication that the ONLY cogroup object of the category of sets is actually the emptyset. $\endgroup$ – Jose Capco Nov 2 '09 at 22:46

$\begingroup$ @Critch: Perhaps. I'm used to thinking of objects in monoidal categories, where the monoidal structure is given beforehand. Because in an abelian category, for instance, since product=coproduct, there are no nontrivial groups and no nontrivial cogroups, whereas there are many interesting Hopf algebras. $\endgroup$ – Theo JohnsonFreyd Nov 3 '09 at 2:12
It seems to me that there is still something to say here.
Suppose C is a category. Then we have the Yoneda functor C(X,) from C to Sets.
A cogroup structure on X is a lift of this functor to a functor to Groups. If C has finite coproducts, but NOT otherwise, this is equivalent to a counit map X > 0, where 0 is the initial object, NOT the terminal object as previously stated, a comultiplication map
X > X + X (where + denotes the coproduct)
and a coinverse map X > X, making the usual diagrams commute.
Obviously, there are no cogroups in any category where the empty set is the initial object, such as Set or Top. If you use pointed sets of spaces, you might have a chance.
But the best example is in the category of commutative kalgebras, where k is a field. A cogroup object is then just a Hopf algebra, and these appear all over mathematics. You might know them under the name affine group schemes, because a cogroup object in commutative kalgebras is the same thing as a group object in the opposite category of affine schemes over k.
A nice example is GL(n). As a commutative ring this is the polynomial algebra on the n^2 entries in the matrix, with the determinant polynomial inverted. The counit is dual to the identity matrix, so it takes the generator x sub ij to 0 if i is not j, and 1 if it is. The comultiplication is dual to matrix multiplication, so it takes x sub ij to the sum of
x sub ik tensor x sub kj .
The coinverse is dual to the inverse matrix, so x sub ij goes to the formula for the ijentry of the inverse.
Mark
This isn't a homotopy theory answer, but the most widely used type of cogroup in mathematics is an algebraic group. An algebraic group is a cogroup object in the category of commutative rings, or in particular commutative algebras over a field. What is important in this case is that people care a lot about the opposite category, the category of affine schemes, or in the field case affine algebraic varieties. People care just as much about schemes and algebraic varieties in general, but if you want the algebraic groups analogue of connected semisimple Lie groups, then all examples are affine.
In my view, any time you endorse a co(something fancy) object in a category, you are really endorsing the opposite category in general. Of course, at the formal level a category and its opposite category have exactly the same information, but conceptually it's a big leap. For example, noncommutative geometry is largely the study of the opposite category of noncommutative rings. Although in this case, people (correctly) use Hopf algebras as if they were cogroup objects, but they aren't! The tensor product operation on noncommutative algebras is not a coproduct.
Just to round things off, the statement "maps into a group object form a group" means that when we have a group object, say G, in some category then the hom set Hom(X,G) can be given the structure of a group. That is, Hom(X,G) is the underlying set of a group (and this is all natural in X, etc). If you try to make the analogous statement "maps into/out of a cogroup object form a cogroup" then you are looking for a way to make Hom(X,G) or Hom(G,X) into a cogroup. This is either trivial or can't be done, depending on how you interpret "cogroup".
If "cogroup" means "turn the maps for group around" then, as Theo says, this is just "sets with an involution" which isn't all that interesting (at least, if you were expecting a nice theory of cogroups).
If "cogroup" means "turn the category around" then you're asking for cogroup objects in Set (in the standard meaning of "cogroup") which means, in particular, that you want a map from X to the empty set: the terminal object in Set^{op} is the initial object in Set and so a map from the terminal object in Set^{op} to X becomes a map from X to the initial object in Set. So, sadly, the empty set is all you get.
So we take "cogroup object in a category" to mean "group object in the opposite category" because that gives us something interesting. Indeed, once the category gets more structure then cogroup objects become more plentiful than group objects (and likewise for other varieties of algebras).
Just one word of warning. Although a "cogroup object in a category" is a "group object in the opposite category", this is a dangerous way to think of them because a morphism of cogroup objects in a category is not a morphism of group objects in the opposite category. The forgetful functor from cogroup objects goes to the original category, not its opposite. The correct statement is that "The category of cogroup objects in a category is the opposite of the category of group objects in the opposite category.". Or, with hopefully obvious symbols:
DG^{c} = (D^{op}G)^{op}

$\begingroup$ Yes, I should have thought slightly more and pointed out that there are no cogroups (with the coproduct definition) in SET. There are noncounital coassociative comonoids in (SET,+). For a set X, the possible comonoid structures on it are classified by triples (X_L, X_R, f), where X_L and X_R are disjoint subsets of X and f is an arbitrary function from X \setminus (X_L \cup X_R) to X_L \cup X_R. $\endgroup$ – Theo JohnsonFreyd Nov 3 '09 at 2:19
An addition: There is a result saying that there are no nontrivial cogroups in the category of topological spaces  only in the homotopy category...

1$\begingroup$ Indeed, this is quite easy to seeit's impossible to have any comultiplication on a nontrivial pointed set which has a twosided unit. $\endgroup$ – Eric Wofsey Nov 2 '09 at 18:16