nLab
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 $\mathrm{End}({\mathrm{Id}}_x)$ for the set of endo-2-morphisms on $X$. Then:

• horizontal composition and vertical composition define the same monoid object structure on $\mathrm{End}({\mathrm{Id}}_x)$;

• this is an abelian monoid.

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

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

In $\mathrm{End}({\mathrm{Id}}_x)$, 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

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

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

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

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

   so this operation is commutative.

6. Finally,

   $$(ab)c=(ab)(1c)=(a1)(bc)=a(bc),$$(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:

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

  Eckmann-Hilton 1 (video)

  Eckmann-Hilton 2 (video)

The diagram proof is displayed here

• Eugenia Cheng, picture (pdf)

and an animation of it is here

• Andrew Stacey, animation (mpeg)

For higher analogues see within the discussion of commutative algebraic monads at:

• Nikolai Durov, New approach to Arakelov geometry , (arXiv:0704.2030)

• Michael Batanin, The Eckmann-Hilton argument and higher operads , Adv. Math. 217 (2008), no. 1, 334–385; (arXiv:math.CT/0207281)