The Eckmann-Hilton result is as follows: Let $X$ be a set equipped with two binary operations $\circ$ and $\otimes$, and suppose
- $\circ$ and $\otimes$ are both unital, meaning there are identity elements $1_{\circ}$ and $1_{\otimes}$ of $X$ such that $1_\circ \circ a = a = a\circ 1_\circ$ and $1_\otimes \otimes a =a= a \otimes 1_\otimes$ for all $a\in X$.
- $(a\otimes b)\circ (c\otimes d)=(a\circ c)\otimes (b\circ d)$ for all $a,b,c,d\in X$.
Then, $\circ$ and $\otimes$ are the same and in fact commutative and associative.
The theorem statement and proof make sense. However, the Wikipedia article claims that this theorem concerns "two unital magma structures on a set where one is a homomorphism for the other." What does this mean? I know what it means for a two magmas to be homomorphic, but this seems to be saying that each magma structure itself somehow is the homomorphism over the other. I assume this statement relates to condition (2). At first I wondered, I wondered if the statement in question was saying that each operation left-distributes over the other; that $a\otimes (b\circ c)= (a\otimes b)\circ (a\otimes c)$ and $a\circ (b\otimes c)= (a\circ b)\otimes (a\circ c)$ together are equivalent to (2), but I don't think this is true.
What does it mean for two unital magmas to be homomorphisms for each other? Although the Eckmann-Hilton result is probably most known from proofs that higher homotopy groups commute, I am looking for an answer that does not require knowledge of algebraic topology to parse.
Given two magmas $(X, \circ)$ and $(Y, \cdot)$ we have the product magma : $(X \times Y, \odot)$, where the product is given as $$(x_1, y_1) \odot (x_2, y_2) := (x_1 \circ x_2, y_1 \cdot y_2)$$
Now, suppose we have two magmas $(X, \circ)$ and $(X, \otimes)$ with the same underlying set $X$. Then, we can ask whether the map $$f = \otimes : (X, \circ) \times (X,\circ) \to (X,\circ)$$ is as a morphism of magmas. Denote the magma structure on $(X, \circ) \times (X,\circ)$ by $\odot$. Unraveling the definition, we then require for all $a,b,c,d \in X$ $$f\big( (a, b) \odot (c, d)\big) = f(a,b) \circ f(c,d) \Leftrightarrow (a \circ c) \otimes (b\circ d)) = (a \otimes b) \circ (c \otimes d)$$ This is precisely the condition $(2)$.