Why the homotopy groups are groups III

Last time, we defined two functors ${\Omega}$ and ${\Sigma}$ on the category ${\mathbf{PT}}$ of pointed topological spaces and (base-point preserving) homotopy classes of base-point preserving continuous maps. We showed that they were adjoint, i.e. that there was a natural isomorphism

$\displaystyle \hom_{\mathbf{PT}}(X, \Omega Y) \simeq \hom_{\mathbf{PT}}(\Sigma X, Y).$

We also showed that ${\Omega Y}$ is naturally an H group, i.e. a group object in ${\mathbf{PT}}$ , for any ${Y}$ . So, given that we have a group operation ${\Omega Y \times \Omega Y \rightarrow \Omega Y}$ , it follows that ${\hom_{\mathbf{PT}}(X, \Omega Y)}$ is naturally a group for each ${Y}$ .

I should perhaps briefly explain that the category ${\mathbf{PT}}$ admits products. The product of ${(X, x_0), (Y, y_0)}$ is just ${(X \times Y, (x_0, y_0))}$ . It also admits coproducts: the coproduct of ${(X, x_0), (Y, y_0)}$ is the wedge sum ${X \vee Y}$ of ${X \sqcup Y}$ with the points ${x_0, y_0}$ identified to one (base)point. (This really should have been mentioned at the beginning. Otherwise, the notion of a group or cogroup object doesn’t make sense.)

It follows that the functor

$\displaystyle Y \rightarrow \hom_{\mathbf{PT}}(\Sigma X, Y)$

can be viewed as taking values in the category of groups. Thus, ${\Sigma X}$ must be, by Yoneda nonsense, a cogroup object in ${\mathbf{PT}}$ , or what we called an H cogroup. It can be checked that the cogroup structure on ${\Sigma X}$ , up to homotopy, of the form

$\displaystyle \Sigma X \rightarrow \Sigma X \vee \Sigma X$

that sends the path ${(x,t)}$ in ${\Sigma X}$ to the path ${(x, 2t) }$ in the first copy of ${\Sigma X}$ (in ${\Sigma X \vee \Sigma X}$ ) followed by ${(x, 2t-1)}$ in the second copy.

0.1. Homotopy groups

Let ${(S^n, s_0)}$ be the ${n}$ -sphere with some basepoint. Then I claim that ${\Sigma(S^n)}$ is just ${S^{n+1}}$ (with some basepoint).

The easiest way for me to see this is to think of ${I^n/\partial I^n}$ (with the basepoint being ${\partial I^n}$ ) instead of ${S^n}$ . Indeed, we have a homeomorphism

$\displaystyle I^n / \partial I^n \simeq S^n.$

When we suspend ${I^n/\partial I^n}$ , we take the product ${I \times I^n}$ , and then collapse the two ends ${\left\{0\right\} \times I^n, \left\{1\right\} \times I^n}$ to points. We then collapse the entire product ${I \times \partial I^n}$ and collapse that to a point. The total equivalence relation that we end up quotienting out by is just that which identifies all of ${\partial I^{n+1} = I \times \partial I^n \cup \left\{0\right\} \times I^n \cup \left\{1\right\} \times I^n}$ to a point.

So it is clear that the suspension of ${(I^n/\partial I^n, \partial I^n)}$ is ${I^{n+1}, \partial I^{n+1}}$ . This implies the claim about ${S^n}$ . In particular:

Proposition 1 ${S^n}$ is an H cogroup.

So the sets ${\pi_n(X, x_0) = \hom_{\mathbf{PT}}(S^n, X)}$ of homotopy classes of continuous maps are actually groups, because ${S^n}$ is an H cogroup, in turn because ${S^n}$ is a suspension.

1. ${\pi_2}$ is abelian

There is a very good reason for the higher homotopy groups to be abelian. A visual illustration of this, which I recommend looking at, is given in Hatcher’s Algebraic Topology. I want to discuss the categorical picture.

Fix ${n \geq 2}$ . We know that suspension and loop space are adjoint functors on the category ${\mathbf{PT}}$ in which we are currently interested. In particular, we have that

$\displaystyle \pi_n(X, x_0) = \hom_{\mathbf{PT}}(S^n, X) = \hom_{\mathbf{PT}}(\Sigma(S^{n-2}), \Sigma X)$

by the adjoint property.

The point is that this has two natural group structures. On the one hand, ${S^{n-1} = \Sigma S^{n-2}}$ is an H cogroup. On the other hand, ${\Sigma X}$ is an H group by catenation of paths.

We now appeal to a general principle to show that ${\pi_n}$ is abelian.

Proposition 2 Let ${X}$ be an H group and ${Y}$ an H cogroup. Then the two group structures on $\displaystyle \hom_{\mathbf{PT}}(Y, X)$

are equal, and ${\hom_{\mathbf{PT}}(X,Y)}$ is an abelian group under this structure.

Proof: The idea is that we have two “independent” ways of defining the group law on ${X, Y}$ . Namely, one has the “unit” for the two group laws: it is just the constant map ${X \rightarrow Y}$ whose image is the basepoint of ${Y}$ . Moreover, the independence statement states that the two operations distribute over each other. This is intuitively clear because one operation depends on the action of ${X}$ and one operation depends on the action of ${Y}$ . But it should be formally checked.

Now, the Eckmann-Hilton argument implies that the two group operations are the same and abelian. $\Box$

2. The Eckmann-Hilton argument

We will prove this more general fact from the following even more general argument:

Proposition 3 (Eckmann-Hilton argument) Suppose ${X}$ is a set and ${\ast, \cdot}$ are two operations ${X \times X \rightarrow X}$ which have a common identity element ${e \in X}$ and which are mutually distributive, i.e. $\displaystyle (x \cdot y) \ast (x' \cdot y') = (x \ast y) \cdot (x' \ast y').$

Then ${\ast, \cdot}$ are equal, and the operation is commutative.

The proof of the Eckmann-Hilton argument is essentially formal manipulation. We write:

$\displaystyle x \ast y = ( x \cdot e ) \ast (e \cdot y)= (x \ast e) \cdot (e \ast y) = x \cdot y.$

Hence, from that formal manipulation, ${\ast, \cdot}$ are equal. Now let us prove commutativity:

$\displaystyle x \ast y = (e \cdot x) \ast (y \cdot e) = (e \ast y) \cdot (x \ast e) = y \cdot x = y \ast x$

by what has already been proved.

Climbing Mount Bourbaki