The Eckmann–Hilton Argument

We describe the Eckmann–Hilton argument and apply it to topological groups.

I recently had occasion to review the Eckmann–Hilton argument for my algebraic topology class. Here is the most general version of this statement.

Theorem 1 (Eckmann–Hilton). Let \(M\) be a set equipped with two operations \(\cdot_1\) and \(\cdot_2\) which have units \(1_1\) and \(1_2\). Suppose further that we have the distributive law \[(a\cdot_1b)\cdot_2(c\cdot_2d)=(a\cdot_1c)\cdot_2(a\cdot_1d)\] for any \(a,b,c,d\in M\). Then \(1_1=1_2\), \({\cdot_1}={\cdot_2}\), and the operations are associative and commutative.

Proof. This is purely formal. For each claim, the main point is to figure out how to apply the given identity.

• We show that \(1_1=1_2\). Well, \[1_1=1_1\cdot_11_1=(1_2\cdot_21_1)\cdot_1(1_1\cdot_21_2)\stackrel*=(1_2\cdot_11_1)\cdot_2(1_1\cdot_11_2)=1_2\cdot_21_2=1_2.\] In the sequel, we will denote the unit by \(1=1_1=1_2\).

• We show that \({\cdot_1}={\cdot_2}\). Well, for any \(a\) and \(b\), we see \[a\cdot_1b=(a\cdot_21)\cdot_1(1\cdot_2b)\stackrel*=(a\cdot_11)\cdot_2(1\cdot_1b)=a\cdot_2b.\] In the sequel, we will denote the operation by \({\cdot}={\cdot_1}={\cdot_2}\).

• We show that \(\cdot\) is commutative. Well, for any \(a\) and \(b\), we see \[a\cdot b=(1\cdot a)\cdot(b\cdot1)\stackrel*=(1\cdot b)\cdot(a\cdot1)=b\cdot a.\]

• We show that \(\cdot\) is associative. Well, for any \(a\), \(b\), and \(c\), we see \[a\cdot(b\cdot c)=(a\cdot1)\cdot(b\cdot c)=(a\cdot b)\cdot(1\cdot c)=(a\cdot b)\cdot c.\]

This completes the proof. \(\blacksquare\)

Here is a particular cute application: group objects it the category \(\mathrm{Grp}\) are just abelian groups.

Definition 2. Fix a category \(\mathcal C\) which admits finite products, and let the terminal object be \(1\). Then a group object \(G\) in \(\mathcal C\) is an object \(G\) equipped with morphisms \(m\colon G\times G\to G\) and \(i\colon G\to G\) and \(e\colon1\to G\) satisfying the following.

• Associativity: the maps \(G\times G\times G\to G\) defined by \(m\circ(\mathrm{id}_G,m)\) and \(m\circ(m,\mathrm{id}_G)\) are equal.
• Identity: the maps \(G\to G\) defined by \(m\circ(\mathrm{id}_G,e)\) and \(m\circ(e,\mathrm{id}_G)\) are both equal to the identity \(\mathrm{id}_G\).
• Inverses: the maps \(G\to G\) defined by \(m\circ(\mathrm{id}_G,i)\) and \(m\circ(i,\mathrm{id}_G)\) are both equal to the map \(G\to1\stackrel e\to G\).

Proposition 3. Let \(G\) be a group object in the category \(\mathrm{Grp}\). Then the map \(m\colon G\times G\to G\) is the usual group multiplication, and \(G\) is abelian.

Proof. We have two binary operations on \(G\): the usual group operation \(\cdot\) and the multiplication \(m\colon G\times G\to G\). The former is unital automatically, and the latter is unital (by the identity axiom) with unit given by the image of \(e\colon1\to G\).

Now, to apply the Eckmann–Hilton argument, we note that \(m\colon G\times G\to G\) is required to be a group homomorphism, which means that \[m(ab,cd)=m((a,c)\cdot(b,d))=m(a,c)m(b,d).\] This is exactly the identity needed for the Eckmann–Hilton argument, so we are done! \(\blacksquare\)

Let’s apply to this a little algebraic topology: we work with the fundamental group of topological groups.

Proposition 4. Let \(X\) be a topological group with identity \(e\). Then \(\pi_1(X,e)\) is an abelian group.

Lemma 5. The functor \(\pi_1\colon\mathrm{Top}_*\to\mathrm{Grp}\) preserves finite products.

Proof. By indution, it is enough to show that \(\pi_1\) preserves the terminal object and preserves products of two pointed spaces. Preserving the terminal object amounts to showing that \(\pi_1(\mathrm{pt},\mathrm{pt})\) is trivial, which is true because the only loops \([0,1]\to\mathrm{pt}\) are already forced to be constant.

It remains to show that \(\pi_1\) preserves the product of two pointed spaces. In other words, for any two pointed spaces \((X,x)\) and \((Y,y)\), we need to show that the projections induce an isomorphism \[\pi_1(X\times Y,(x,y))\to\pi_1(X,x)\times\pi_1(Y,y).\] This is certainly a group homomorphism, so we just need to check that it is a bijection.

• Surjective: two loops \(\gamma_X\colon[0,1]\to X\) and \(\gamma_Y\colon[0,1]\to Y\) can be placed into a single loop \((\gamma_X,\gamma_Y)\colon[0,1]\to X\times Y\) which projects down correctly to \(\gamma_X\) and \(\gamma_Y\).
• Injective: any loop \(\gamma\colon[0,1]\to X\times Y\) can be written in the form \((\gamma_X,\gamma_Y)\) where \(\gamma_X\) and \(\gamma_Y\) are the two projections down to \(X\) and \(Y\), respectively. If \(\gamma_X\) and \(\gamma_Y\) are both homotopic to the constant paths, then we can use the two homotopies coordinate-wise to show that \(\gamma\) is homotopic to the constant path as well. \(\blacksquare\)

Proof of Proposition 4. We use Proposition 3, which tells us that it is enough to check that \(\pi_1(X,e)\) is a group object in \(\mathrm{Grp}\). An equivalent way to say that \(X\) is a topological group is to say that \(X\) is a group object in \(\mathrm{Top}\) or even \(\mathrm{Top}_*\) where the identity is taken to be the basepoint.

We are ready to show that \(\pi_1(X,e)\) is a group object in \(\mathrm{Grp}\). This more or less follows from Lemma 5. Indeed, being a group object amounts to having certain morphisms on products and then checking that certain maps constructed from these (via compositions and products) are equal. All these data are preserved by a functor which preserves products, so we conclude that \(\pi_1(X,e)\) is in fact a group object. \(\blacksquare\)