Eckmann-Hilton and higher homotopy groups

Solution 1:

The essential point is the following (I'll only talk about $\pi_2$ but everything I say here generalizes). Let $H : I^2 \to X$ be a continuous map, where $I = [0, 1]$ is the interval (so that $I^2$ is a square). Then $H$ may be regarded as a homotopy between two paths $I \to X$ in two ways:

• as a homotopy between the paths $H(0, -)$ and $H(1, -)$, or
• as a homotopy between the paths $H(-, 0)$ and $H(-, 1)$.

This induces two notions of composition called vertical and horizontal composition on such maps, one of which stacks the corresponding squares vertically (when the corresponding paths agree) and one of which stacks the squares horizontally. Both of these operations are associative up to homotopy, and the condition that each of them is a homomorphism for the other (up to homotopy) is precisely the statement that you can interpret the mushing of four squares together into a single big square in two different ways (stacking horizontally then vertically, or stacking vertically then horizontally) and these are the same thing. (This can be thought of as a $2$-dimensional statement in the vein of of the associativity of composition of functions, which is a $1$-dimensional statement about stacking lines.)

Now add the additional condition that all four sides of the square are mapped to the basepoint $x_0$. Then there is an identity square $i$ (for both vertical and horizontal composition) which sends everything to $x_0$, and running the Eckmann-Hilton argument here amounts to pushing squares around. In ASCII art form, it looks like this:

-----   ---------               ---------   ----- 
| f |   | i | f |   ---------   | g | i |   | g |
----- = --------- = | g | f | = --------- = -----
| g |   | g | i |   ---------   | i | f |   | f |
-----   ---------               ---------   -----

The above is really an argument in the $2$-category $\Pi_2(X)$. See my blog post Centers, 2-categories, and the Eckmann-Hilton argument for an elaboration on this. In the notation of that blog post the Eckmann-Hilton argument looks like this:

Solution 2:

Recall that the Eckmann-Hilton argument says that if you have two different unital monoid structures on a set such that both are homomorphisms for the other and both have the same unit, then they're equal and commutative. We want to apply that to see that higher homotopy groups are commutative.

The two different monoid structures we want to consider are 'vertical' and 'horizontal' concatenation of maps from $I^n$: if I have two maps $f,g:I^n\to X$ for some topological space $X$ so that $f(\partial I^n)=g(\partial I^n)=x_0$, then I can concatenate them in the following ways as long as $n>1$:

$f\cdot g=\begin{cases} f(2x,y,\cdots) &\text{ for } 0\leq x\leq 1/2\\ g(2x-1,y,\cdots) &\text{ for } 1/2\leq x\leq 1 \end{cases}$

$f \star g= \begin{cases} f(x,2y,\cdots) &\text{ for } 0\leq y\leq 1/2\\ g(x,2y-1,\cdots) &\text{ for } 1/2\leq y\leq 1 \end{cases}$

Now, if I can show that these are homomorphisms for each other, then I'll have that $\pi_n$ is commutative for $n>1$. But this is easy: both $(f\cdot g)\star (h\cdot i)$ and $(f\star h)\cdot (g\star i)$ are the following map:

$(x,y,...)\mapsto\begin{cases} f(2x,2y,\cdots) &\text{ for }(x,y)\in [0,1/2]\times [0,1/2]\\ g(2x-1,2y,\cdots) &\text{ for } (x,y)\in [1/2,1]\times [0,1/2]\\ h(2x,2y-1,\cdots) &\text{ for } (x,y)\in [0,1/2]\times [1/2,1]\\ i(2x-1,2y-1,\cdots) &\text{ for } (x,y)\in [1/2,1]\times [1/2,1] \end{cases}$

Now, by the Eckmann-Hilton argument, these two sorts of composition are the same and are in fact commutative.