# nLab Eckmann-Hilton argument

### Context

#### Higher category theory

higher category theory

# The Eckmann–Hilton argument

## Statements

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:

###### Proposition

Let $C$ be a 2-category and $x\in C$ an object. Write ${\mathrm{Id}}_{x}$ for the identity morphism of $X$ and $\mathrm{End}\left({\mathrm{Id}}_{x}\right)$ for the set of endo-2-morphisms on $X$. Then:

On the face of it, this is a special case of the general situation, although in fact every case is an example for appropriate $C$.

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:

1. the other is also a homomorphism with respect to the first;
2. each also preserves the other's identity;
3. the identities are the same;
4. the operations are the same;
5. the operation is commutative;
6. the operation is associative.

This can also be internalised in any monoidal category.

## Proofs

A pasting diagram-proof of 1 is depicted in Cheng below. Here we prove the $6$-element general form in $\mathrm{Set}$.

###### Proof

The basic equation that we have (that one operation $*$ is a homomorphism with respect to another operation $\circ$) is

$\left(a\circ b\right)*\left(c\circ d\right)=\left(a*c\right)\circ \left(b*d\right).$(a \circ b) * (c \circ d) = (a * c) \circ (b * d) .

In $\mathrm{End}\left({\mathrm{Id}}_{x}\right)$, this is the exchange law.

We prove the list of results from above in order:

1. Simply read the basic equation backwards to see that $\circ$ is a homomorphism with respect to $*$.

2. Now if ${1}_{*}$ is the identity of $*$ and ${1}_{\circ }$ is the identity of $\circ$, we have

${1}_{\star }\circ {1}_{\star }=\left({1}_{\star }\circ {1}_{\star }\right)*{1}_{\star }=\left({1}_{\star }\circ {1}_{\star }\right)*\left({1}_{\star }\circ {1}_{\circ }\right)=\left({1}_{\star }*{1}_{\star }\right)\circ \left({1}_{\star }*{1}_{\circ }\right)={1}_{\star }\circ {1}_{\circ }={1}_{\star }.$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.

3. Then

${1}_{\star }={1}_{\star }*{1}_{\star }=\left({1}_{\star }\circ {1}_{\circ }\right)*\left({1}_{\circ }\circ {1}_{\star }\right)=\left({1}_{\star }*{1}_{\circ }\right)\circ \left({1}_{\circ }*{1}_{\star }\right)={1}_{\circ }\circ {1}_{\circ }={1}_{\circ },$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 $1$.

4. Now

$a*b=\left(a\circ 1\right)*\left(1\circ b\right)=\left(a*1\right)\circ \left(1*b\right)=a\circ b,$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.

5. Then

$ab=\left(1a\right)\left(b1\right)=\left(1b\right)\left(a1\right)=ba,$a b = (1 a) (b 1) = (1 b) (a 1) = b a ,

so this operation is commutative.

6. Finally,

$\left(ab\right)c=\left(ab\right)\left(1c\right)=\left(a1\right)\left(bc\right)=a\left(bc\right),$(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 $\mathrm{Mon}$, 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:

$a*b=\left(a\circ 1\right)*\left(1\circ b\right)=\left(a*1\right)\circ \left(1*b\right)=a\circ b=\left(1*a\right)\circ \left(b*1\right)=\left(1*b\right)\circ \left(a*1\right)=b*a,$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.

## Corollaries

A $2$-tuply monoidal $0$-category, if defined as a pointed simply connected bicategory, is also the same as an abelian monoid.

A $2$-tuply monoidal $1$-category, if defined as a pointed simply connected tricategory, is the same as a braided monoidal category.

Every homotopy group ${\pi }_{n}$ for $n\ge 2$ is abelian.

## History

The beautiful and powerful Eckmann-Hilton argument is due to Beno Eckmann and Peter Hilton.

## References

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:

Revised on November 11, 2010 02:55:17 by Toby Bartels (76.85.201.22)