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.
Best Answer
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.