n-category = (n,n)-category
n-groupoid = (n,0)-category
In its usual form, the Eckmann–Hilton argument shows that a monoid or group object in the category of monoids or groups is commutative. In other terms, if a set is equipped with two monoid structures, such that one is a homomorphism for the other, then the two structures coincide and the resulting monoid is commutative.
From the nPOV, we may want to think of the statement in this way:
On the face of it, this is a special case of the general situation, although in fact every case is an example for appropriate .
A more general version is this: If a set is equipped with two binary operations with identity elements, as long as they commute with each other in the sense that one is (with respect to the other) a homomorphism of sets with binary operations, then everything else follows:
The basic equation that we have (that one operation is a homomorphism with respect to another operation ) is
(a \circ b) * (c \circ d) = (a * c) \circ (b * d) .
In , this is the exchange law.
We prove the list of results from above in order:
Simply read the basic equation backwards to see that is a homomorphism with respect to .
Now if is the identity of and is the identity of , we have
1_\star \circ 1_\star = (1_\star \circ 1_\star) * 1_\star = (1_\star \circ 1_\star) * (1_\star \circ 1_\circ) = (1_\star * 1_\star) \circ (1_\star * 1_\circ) = 1_\star \circ 1_\circ = 1_\star .
A similar argument proves the other half.
1_\star = 1_\star * 1_\star = (1_\star \circ 1_\circ) * (1_\circ \circ 1_\star) = (1_\star * 1_\circ) \circ (1_\circ * 1_\star) = 1_\circ \circ 1_\circ = 1_\circ ,
so the identities are the same; we will now write this identity simply as .
a * b = (a \circ 1) * (1 \circ b) = (a * 1) \circ (1 * b) = a \circ b ,
so the operations are the same; we will write them both with concatenation.
a b = (1 a) (b 1) = (1 b) (a 1) = b a ,
so this operation is commutative.
(a b) c = (a b) (1 c) = (a 1) (b c) = a (b c) ,
so the operation is associative.
If you start with a monoid object in , then only (4&5) need to be shown; the others are part of the hypothesis. This classic form of the Eckmann–Hilton argument may be combined into a single calculation:
a * b = (a \circ 1) * (1 \circ b) = (a * 1) \circ (1 * b) = a \circ b = (1 * a) \circ (b * 1) = (1 * b) \circ (a * 1) = b * a ,
where the desired results involve the first, middle, and last expressions.
An expositions of the argument is given here:
The diagram proof is displayed here
and an animation of it is here
For higher analogues see within the discussion of commutative algebraic monads at: