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.

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