I’ve been reading a lot of algebraic topology as of late, but I still don’t think I understand most of it properly. I will try to blog about some of the ideas as a means to understand the ideas better.

The **homotopy groups** are essentially homotopy classes of maps from the sphere into a space . For instance, the first homotopy group (or **fundamental group**) is the group of homotopy classes of loops at a fixed base-point. But why should they be a group? There is a categorical reason for this, and while it’s not immensely deep, I’d like to explain it.

**1. Definition **

Fix a topological space with a basepoint . Let . We consider the set of homotopy classes of maps from the unit -cube into . This, however, is not a very interesting set, because is contractible; there is thus only one element. Instead, we consider the set of homotopy classes of maps that send the boundary into . When I say “homotopy classes,” I mean that homotopies are required to send the boundary into .

To make this explicit:

Definition 1The-th homotopy groupof a space-with-basepoint is the set of all maps sending to under the equivalence relation that are considered equivalent if there exists a homotopy

between relative to .This is functorial in the obvious way: a map (i.e. a map sending ) induces a map on the homotopy groups.

So this is a set. But we have not explained why this is supposed to be a homotopy *group*. The group structure can be visualized intuitively in the case : given that each send the boundaries into , we can **catenate** to get a path

This multiplication is not associative on the loops, but it is associative up to homotopy. And that’s all you need to define the group law on the set of homotopy classes when . A modification of this works in higher dimensions.

The essential point here was that paths can be pieced together back to back. This is an essential point of the space . For instance, it would be less obvious how to make the set of homotopy classes from a pointed torus into a space a group in any natural way.

The answer to the earlier question is that (and the higher ) is “almost” a co-group object in the category of pointed topological spaces. I say almost, because the comultiplication is (co)associative only up to homotopy. In particular, the fact that all these homotopy groups are groups is a special case of Yoneda categorical nonsense.

**2. The homotopy category **

Since we work with maps up to homotopy to define the groups, it makes sense to consider a corresponding category. In this category, the functors are representable by the objects . Thus the fact that they can be viewed naturally as groups is a co-group statement about the .

Definition 2We consider the category , called the homotopy category of pointed topological spaces, defined as follows.The objects of are pointed spaces . So is a topological space and .

The morphisms between are the set of maps with with the equivalence relation that are equivalent if there is a homotopy between such that .

The last condition says that is a homotopy **relative to **. Strictly speaking, one should check something first. Namely, that if are equivalent and is a pointed map, then are equivalent; this is necessary to see that the category is well-defined. But this is straightforward.

So what is the set in this language? Recall that a map sending is the same thing as a map out of the quotient space sending the “basepoint” . So if we regard as a pointed space, then is just

It is clear that is a covariant functor from . It is also clear that it is co-representable by the object . But in fact, as asserted above, each is a group.

Proposition 3Let be any category with finite coproducts and a final object. Suppose and the sets are endowed with a group structure which is natural. In other words, can be viewed as a functor to the category of groups.Then is a co-group object.

Let us recall what it means for to be a co-group object. In other words, there is a **comultiplication **map

and a **counit map**

for the final object. These are required to satisfy arrow-theoretic reversals of the usual relations imposed on a group. For instance, the map

composing the counit and the comultiplication, as to be the identity.

Now has a final object: it’s just , a one-point space. Moreover, it has finite coproducts. The coproduct of is the space . This is not completely trivial, actually. On the one hand, it is clear that to give a pointed map of topological spaces

is the same as giving a pair of pointed maps . We need to check, however, that a pair of homotopies

induce

The reason is that piece together to give a map of the quotient space of and with for each . By the following theorem in general topology, this quotient space is the same as the product . So the homotopies indeed do patch appropriately.

Lemma 4Let be a quotient map. If is a locally compact Hausdorff space, then is a quotient map.

We won’t prove this nontrivial but not terribly difficult result from general topology.

So, let’s put this together. has finite coproducts and a final object. So when a (co)representable functor has its image in the category of groups, then is a cogroup in . Unravelling the definitions shows that this means that there are continuous maps of pointed topological spaces

and

such that coassociativity holds *up to pointed homotopy*.

Definition 5AH cogroupis a pointed topological space with comultiplication and counit maps such that the cogroup diagrams commute up to pointed homotopy. In other words, a cogroup object in .

So we see that necessarily is an cogroup, which is what lets one catenate a pair of maps . This cogroup structure in the case , for instance, sends to in the first copy of if and to in the second copy of for . Since the boundaries of the two are identified, this is a well-defined comulitiplication map, and one can check that it is homotopy co-associatve. The counit collapses everything to a point.

And that is one basic reason for why the homotopy groups should be groups.

This still doesn’t explain why we would expect to be an cogroup. Next, I will discuss a pair of adjoint functors on that helps answer this related question, as well as the further question of why the higher homotopy groups are commutative.

September 26, 2010 at 11:38 pm

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