How should you think about cogroups intuitively?

Question

There have been various previous questions about cogroup objects on MSE and MathOverflow, mainly focussing on why various examples are indeed cogroup objects (e.g. spheres are cogroup objects in the category $\text{hTop}_\bullet$ of pointed topological spaces up to homotopy, abelian groups are cogroup objects in the category of groups etc.). None have focused on intuitively what cogroups are.

When we think about groups, one natural way of thinking about the definition is that it kind of axiomatises the symmetries of some object. This on its own has various facets:

• You can use this to directly come up with heuristic justifications for the axioms and why extra axioms aren't necessary (e.g. symmetries don't in general commute, so groups shouldn't in general commute).
• It suggests that there should be a meaningful interplay between a group $G$'s structure and its group actions or $G$-sets (which is of course very much true).
• It can even be partially encoded as a theorem, i.e. Cayley's Theorem showing that every group is specifically a permutation group.

What is an analogous perspective on cogroups? There are many aspects I find mysterious... To name a couple:

• I struggle to understand what on earth the unary operation $C\to C\sqcup C$ on a cogroup $C$ could represent. It takes an element... to something in the coproduct of $C$ with itself? So a little like the disjoint union of $C$ with itself?! What then does coassociativity of this operation really mean?!
• The only cogroup object in the category $\text{Set}$ is the empty set! At least in the case of groups, group objects in $\text{Set}$ recover the original actual notion of groups...

Ideally an answer will provide a nice conceptual basis on which to understand cogroups as described above.

Answer 1

I think that for any question of the form "how should I think of X intuitively", its hard not to write a long rambly, blog post of an answer. Anyways, I'll try to keep this based on concrete examples. A good way to understand cogroups is in the context of how they relate to other coalgebraic structures, and how the dual algebraic structures relate to each other. A good example to look at is categories and cocategories. There are simple examples of each and categories have rich relations to other algebraic objects that allows us to see how tinkering with something out the coalgebraic side affects things on the algebraic side. For a longer discussion, you can look for resources on internal categories and cocategories.

I'll give an abridged definition of a cocategory. Fix an ambient category $C$ with pushouts. For now we'll think of a cocategory in $C$ like a directed interval that is equipped with a notion of gluing two copies of itself tip to tail. More explicitly, a cocategory in $C$ consists of two objects $c_0$ and $c_1$, which we think of as the point object and the arrow object, respectively. These come equipped with maps $\delta_0 , \delta_1 : c_0 \to c_1$, which correspond to the inclusions of the point to the ends of the arrow. Then the push out

$$ c_1 \xleftarrow{\delta_1} c_0 \xrightarrow {\delta_0} c_1 $$

is two copies of $c_1$ glued to each other codomain to domain. A cocategory comes equipped with a ``cocomposition'' map $c_1 \to c_1 \sqcup_{c_0} c_1$. This corresponds to an arrow in the pushout with the domain of the left arrow and codomain of the right arrow. So it goes all the way across both copies of the original arrow. There is other structure required that I won't mention here.

An example of a cocategory can be found in $\mathsf{Cat}$. It is given by the categories ${\bf 1}$ and ${\bf 2}$ with the natural inclusions ${\bf 1} \to {\bf 2}$. The cocomposition map is the map ${\bf 2} \to {\bf 3}$ that sends the arrow in ${\bf 2}$ to the composite arrow in ${\bf 3}$. If we drop the cocomposition map from the definition, we basically have a 1-truncated cosimplical object. So there is a sense in which a cocategory is a geometric gadget that records a (possibly directed) interval with a notion of gluing two copies of it together in a way that results in another arrow.

To relate this back to categories in $\mathsf{Set}$, we can use this any cocategory to define an ordinary category. Suppose $c_0$ and $c_1$ are a cocategory as above and fix $x$ in $C$. We define the set of arrows to be $C(c_1,x)$. Then can define precomposition by the inclusions $\delta_0^\ast , \delta_1^\star : C(c_1 , x) \to C(c_0 , x)$ to be the domain and codomain functions. We can show that composable arrows are naturally equivalent to maps out of $c_1 \sqcup_{c_0} c_1$, so precomposition with comultilpication gives a function from composable arrows to arrows.

We can put restrictions on our cocategory structure to obtain related coalgebraic structures. If the point object $c_0$ is initial in $C$, then the arrow has a unique point and is thus a (directed) loop. This is defines something called a comonoid. There is a natural comonoid structure on the "directed circle" (a category freely generated by a single loop) in the category of pointed categories. A comonoid naturally gives rise to an algebraic structure by precomposing with the comonoid structure. In this case, maps out of $c_1 \sqcup_{c_0} c_1$ are equivalent to pairs of maps out of $c_1$. Thus every arrow is composable, so we indeed return the notion of a monoid.

We can also ask that the arrow $c_1$ can be "turned around". This gives rise to a cogroupoid. Sometimes these are also called interval objects since they behave like undirected intervals. An important examples is the topological interval $[0,1]$ in the homotopy category. The algebraic structure induced by precomposition with this cogroupoid is the fundamental groupoid of a space. Another example is the walking isomorpism category $I$. We can define the underlying groupoid, or core of a category, using the cogroupoid $I$.

If we require both that $c_0$ is initial and $c_1$ can be turned around, then we get the concept of a cogroup. The most fundamental examples of cogroups are the spheres $S^n$. The cogroup structure on $S^n$ has a structure map $S^n \to S^n \lor S^n$, which corresponds to going around the left copy, then going around the right copy. Up to homotopy, this forms a cogroup. Precomposing by this cogroup structure is the definition the algebraic notion of homotopy group $\pi_n$.

Of course this is just a heuristic. Formally speaking, every cogroup is a group. And others have already mentioned algebraic notions of cogroups in the comments. I just think these are good motivating examples. I have a half lucid thought relating this to the some nlab article. But I won't ramble here...

Answer 2

Kevin alluded to this, but I think the quickest path to an "intuitive" understanding of cogroups is via their universal property. If $G$ is a group, then $\text{Hom}_\mathsf{Set}(X,G)$ is a group for any set $X$. Indeed, given functions $f,g : X \to G$ we can define

• $\underline{1} = x \mapsto 1_G$
• $fg = x \mapsto f(x) \cdot_G g(x)$
• $f^{-1} = x \mapsto f(x)^{-1}$.

More generally, in any category, if $G$ is a group object in $\mathcal{C}$, then $\text{Hom}_\mathcal{C}(X,G)$ is a group for any object $X$! This is true for basically the same reason, and it makes a good exercise if you haven't checked this.¹

Now if we put our category theorist hat on, we might want to dualize this. After all, the fact that $\text{Hom}_\mathsf{Set}(X,G)$ is always a group comes in handy pretty often! Is there a kind of object $H$ so that $\text{Hom}_\mathcal{C}(H,X)$ is always a group? The answer is yes, and this is precisely the definition of a cogroup²!

So the first intuitive reason to care about cogroups is that we don't care about cogroups! We care about objects that give us families of groups! For instance, the first reason to care that $S^1$ is a cogroup isn't because it tells you something interesting about $S^1$ directly. It's because it tells yous that $\text{Hom}_{\text{Ho}(\mathsf{Top}_*)}(S^1,X)$ is always a group! Indeed, it's $\pi_1(X)$! A cogroup $H$ is a super efficient way of getting a family of interesting groups $\text{Hom}_\mathcal{C}(H,X)$ varying naturally in $X$! And that's the interesting thing! (For instance, it gives us a nice algebraic invariant of $X$)

Another reason to care is because we care about opposite categories. Many "geometric" categories are defined to be the opposite of some well known "algebraic" category. For instance, the category of affine schemes is defined to be the opposite of the category of (commutative) rings. Or the category of locales is defined to be the opposite of the category of frames. Now we're naturally interested in affine group schemes or a localic groups (which are like lie groups and topological groups, respectively), and a natural question is whether we can understand these purely in terms of the algebraic structure on the opposite side of the duality (since this is the side where we do most of our computations). The answer is "yes", since a group in $\mathsf{Aff} = \mathsf{Ring}^\text{op}$ is exactly the same as a cogroup in $\mathsf{Aff}^\text{op} = \mathsf{Ring}$! So again, it's not the cogroup structure itself that we're interested in. It's the fact that it gives us a group structure on the geometric side of the duality!

Of course, once you spend a lot of time with these objects, you come to appreciate them on their own terms, and you start being interested in cogroups for their own sake. I don't think that anybody would have come up with them in the first place, though, were it not for concrete applications (related to groups) like those above.

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I hope this helps ^_^

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¹: In fact there's slightly more to say here. These groups $\text{Hom}_\mathcal{C}(X,G)$ vary naturally in $X$, and you can check that $G$ is a group object in $\mathcal{C}$ if and only if $\text{Hom}_\mathcal{C}(-,G) : \mathcal{C}^\text{op} \to \mathsf{Set}$ is actually valued in groups! There's a few quick ways to see this, but one nice option is to use lawvere theories and the yoneda lemma.

²: This builds on the last footnote. In the same way that "$\text{Hom}_\mathcal{C}(-,G)$ is group valued" isn't the usual definiton of a group object in $\mathcal{C}$ (instead we usually have arrows and axioms on those arrows), the definition $\text{Hom}_\mathcal{C}(H,-)$ is group valued" isn't the usual definition of a cogroup object. Of course, dualizing the exercise from footnote $1$ will give a proof that this agrees with the usual definition of a cogroup object in terms of arrows and axioms.