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Let $[X, Y]_0$ denote base point preserving homotopy classes of maps $X\rightarrow Y$. A multiplication on a pointed space $Y$ is a map $\phi: Y\times Y\rightarrow Y.$ From this map, we can define a continuous map for each pointed space $X$, $\phi_X: [X, Y]_0\times [X, Y]_0\rightarrow [X, Y]_0,$ by the composition $$\phi_X (\alpha, \beta)(x)=\phi(\alpha(x), \beta(x)).$$ If $([X, Y]_0, \phi_X)$ is a group for each $X$, then $(Y, \phi)$ is called a homotopy associative $H$-space.

A $coH$-space is defined from a comultiplication, namely, a map $\psi: X\rightarrow X\vee X.$ Then, for each pointed space $Y$, we can define a function $\psi^Y: [X, Y]_0\times [X, Y]_0\rightarrow [X, Y]_0$ in this way: $$\psi^Y(\alpha, \beta)=(\alpha\vee\beta)\circ\psi.$$ If $([X, Y]_0, \psi^Y)$ is a group for each $Y$, then $(X, \psi)$ is called a homotopy associative $coH$-space.

So, as we can see, if we have a homotopy associative $coH$-space $(X, \psi)$ and a homotopy associative $H$-space $(Y, \phi)$, then we can define two group structures on the space $[X, Y]_0$. My question is: are they "equivalent" in some sense? Obviously, whatever $\phi$ or $\psi$ is, the zero element of the group is the constant map in $[X, Y]_0.$ However, the two group structures do depend on the choice of $\phi$ and $\psi$, which seems have little relationship with each other.

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up vote 8 down vote accepted

I looked at my homotopy theory lecture notes and we had the following similar result: $X$ H-CoGroup, $Y$ $H$-Group, then both group structures defined on [X,Y] agree. The proof goes roughly as follows: Call the upper products $\cdot$, resp. $*$. Inserting the definitions of those products, one can show the following "distributivity":

$(a\cdot b)*(c\cdot d)=(a * c)\cdot(b * d)$

Then one shows that both products have the same neutral element and finally

$f*g=(f\cdot 1) * (1\cdot g)=(f * 1)\cdot(1 * g)=f\cdot g$,

gives the result. That's the strategy of the proof in the case of $H$-(co-)groups.

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Thanks! I really love the "distributivity" equality you gave. It's easy to prove, useful, and elegant. With this formula, we can also prove that both groups are abelian. Thanks! – Megan Feb 28 '10 at 20:35
This is called the Eckmann-Hilton Argument:–Hilton_argument It not only proves that the two products are the same, but also proves that they are commutative and associative (you don't need to assume this!). This argument is also used to prove that $\pi_i$ is an abelian group for $i\geq 2$, since in this case you have at least two multiplications (from gluing spheres in different directions) which satisfy the above distributivity. – Chris Schommer-Pries Feb 28 '10 at 20:44
Hmmm. The above link is broken. Let me try again: <a href="–Hilton_argument">–Hilton_argument</a>. – Chris Schommer-Pries Feb 28 '10 at 20:47
I see how to prove the commutativity of $\pi_i$ with $i\geq 2$. For pointed spaces $X$ and $Y$, let's consider the loop space $\Omega Y$ and the suspension $SX$. Define $\phi: \Omega Y\times\Omega Y\rightarrow\Omega Y$ by $\phi(u,v)=u(2t)$ when $t\in [0,\frac{1}{2}]$ and $=v(2t-1)$ when $t\in [\frac{1}{2},1].$ Then, $(\Omega Y,\phi)$ is a $H$-space. And, define $\psi: SX\rightarrow SX\vee SX$ to be $\psi([x,t])=([x,2t],*)$ when $t\in [0,\frac{1}{2}]$ and $=(*, [x,2t-1])$ when $t\in [\frac{1}{2},1]$. Then, $(SX, \psi)$ is a $coH$-space. – Megan Feb 28 '10 at 21:52
There's also an nlab page on this: and in a talk recently, I came up with an animation showing the key step: – Loop Space Mar 1 '10 at 8:52

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