Eckmann–Hilton argument

Remarks

Put $a = d = 1_\circ$ and $b = c = 1_\otimes$ in the interchange law: the left side is $1_\circ \circ 1_\circ = 1_\circ$ and the right side is $1_\otimes \otimes 1_\otimes = 1_\otimes$, so the units agree; write $1$ for both. Then $a \circ d = (a \otimes 1) \circ (1 \otimes d) = (a \circ 1) \otimes (1 \circ d) = a \otimes d$, so the operations agree. Also $b \circ c = (1 \otimes b) \circ (c \otimes 1) = (1 \circ c) \otimes (b \circ 1) = c \otimes b = c \circ b$, which is commutativity. Associativity follows in the same way from $(a \otimes b) \circ (1 \otimes c) = (a \circ 1) \otimes (b \circ c)$. The units are needed: on a set with at least two elements, the operations $a \circ b = a$ and $a \otimes b = b$ satisfy the interchange law, but they differ. The main application: for $n \ge 2$, the homotopy group $\pi_n(X, x_0)$ carries two compatible products, concatenation in two different coordinates, so it is abelian.

Going deeper

For a student

The argument explains why the higher homotopy groups $\pi_n$ for $n \ge 2$ are abelian, while $\pi_1$ need not be. In dimension $n \ge 2$ there are two independent directions in which to concatenate maps of cubes. For the same reason, the fundamental group of an H-space, such as a topological group, is abelian. There, pointwise multiplication of loops satisfies the interchange law with concatenation. In category theory it shows that a monoid object in the category of monoids is a commutative monoid. It also shows that the endomorphisms of the unit object in a monoidal category form a commutative monoid.