# 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 $Id_x$ for the identity morphism of $X$ and $End(Id_x)$ 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 $Set$.

###### Proof

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

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

In $End(Id_x)$, 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

$1_\star \circ 1_\star = (1_\star \circ 1_\star) * 1_\star = (1_\star \circ 1_\star) * (1_\star \circ 1_\circ) = (1_\star * 1_\star) \circ (1_\star * 1_\circ) = 1_\star \circ 1_\circ = 1_\star .$

A similar argument proves the other half.

3. Then

$1_\star = 1_\star * 1_\star = (1_\star \circ 1_\circ) * (1_\circ \circ 1_\star) = (1_\star * 1_\circ) \circ (1_\circ * 1_\star) = 1_\circ \circ 1_\circ = 1_\circ ,$

so the identities are the same; we will now write this identity simply as $1$.

4. Now

$a * b = (a \circ 1) * (1 \circ b) = (a * 1) \circ (1 * b) = a \circ b ,$

so the operations are the same; we will write them both with concatenation.

5. Then

$a b = (1 a) (b 1) = (1 b) (a 1) = b a ,$

so this operation is commutative.

6. Finally,

$(a b) c = (a b) (1 c) = (a 1) (b c) = a (b c) ,$

so the operation is associative.

If you start with a monoid object in $Mon$, 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:

$a * b = (a \circ 1) * (1 \circ b) = (a * 1) \circ (1 * b) = a \circ b = (1 * a) \circ (b * 1) = (1 * b) \circ (a * 1) = b * a ,$

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 $n \geq 2$ 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 October 19, 2016 12:52:17 by Ingo Blechschmidt (137.250.162.32)