Show that the space $Y = S^3 \vee S^6$ has precisely two distinct homotopy classes of comultiplications. Here is the question:
A comultiplication for a pointed space $X$ is a map $\phi : X \rightarrow X \vee X$ so that the composite $$X \xrightarrow{\phi} X \vee X \xrightarrow{i_{X}} X \times X$$ is homotopic to the diagonal map. Show that the space $X = S^3 \vee S^5$ has infinitely many distinct homotopy classes of multiplications, while the space $Y = S^3 \vee S^6$ has precisely two distinct homotopy classes of comultiplications.
I got this hint:
Use this problem:
(c) Calculate the homotopy fibre of the inclusion $i_{X} : X \vee X \rightarrow X \times X. $
And you probably need the Hilton-Milnor Theorem.
My Question is:
Could anyone help me fill the details please?
 A: Write $X=S^3\vee S^6$ and note that this space admits at least one comultiplication, since it is a suspension. Denote the suspension comultiplication
$$c:X\rightarrow X\vee X$$
and observe that it is coassociative, cocommutative and counital, since $X$ is a double suspension. Thus for each space $Y$, the comultiplication $c$ furnishes the homotopy set $[X,Y]$ with a group structure, which is natural with respect to maps $Y\rightarrow Y'$.
Now recall that the homotopy fibre of the inclusion $j:X\vee X\hookrightarrow X\times X$ is equivalent to $\Sigma \Omega X\wedge \Omega X$ and there is a fibration sequence
$$\dots\Omega(X\vee X)\xrightarrow{\Omega j}\Omega X\times \Omega X\xrightarrow{\delta}\Sigma \Omega X\wedge \Omega X\xrightarrow{w} X\vee X\xrightarrow{j} X\times X$$
which defines the maps $w$ and $\delta$. 
Consider the resulting Puppe sequence
$$\dots\rightarrow [X,\Omega X\times \Omega X]\xrightarrow{\delta_*} [X,\Sigma\Omega X\wedge \Omega X]\xrightarrow{w_*} [X,X\vee X]\xrightarrow{j_*} [X,X\times X],$$
which here is an exact sequence of abelian groups. Since a comultiplication on $X$ is a map $X\rightarrow X\vee X$ which lifts the diagonal $\Delta:X\rightarrow X\times X$ through $j$, all the comultiplications 'live in' $[X,X\vee X]$ and are all mapped down to the same element in $[X,X\times X]$ by $j_*$ .
Thus by exactness we see that the set of comultiplications on $X$ is in bijective correspondence with the image of $w_*$, and so identifies with a certain coset in $[X,\Sigma\Omega X\wedge \Omega X]$. The task now is to identify this coset. Here is the trick: the fibration sequence splits after looping. Indeed, if $pr_i:X\times X\rightarrow X$, $i=1,2$, are the two projections, and $in_i:X\hookrightarrow X\vee X$, $i=1,2$ are the two inclusions, then
$$s=\Omega(in_1 pr_1)+\Omega (in_2pr_2):\Omega (X\times X)\cong \Omega X\times \Omega X\rightarrow \Omega (X\vee X)$$
is a section of $\Omega j$. Here I am using the loop addition on $\Omega (X\vee X)$ to form the sum. 
Now, since $\Omega j$ admits a section, $\delta$ is null-homotopic: $$\delta\simeq \delta(\Omega w)s\simeq (\delta\Omega w)s\simeq \ast s\simeq \ast.$$
In particular, in the Puppe sequence, $\delta_*=0$, and $w_*:[X,\Sigma\Omega X\wedge \Omega X]\rightarrow[X,X\vee X]$ is monic.
The conclusion is that the coset is the entire group, and fixing one comultiplication induces a bijective correspondence between comultiplications on $X$ and homotopy classes of maps $X\rightarrow \Sigma\Omega X\wedge \Omega X$. All of this follows because of the presence of at least one comultiplication on $X$, namely $c$. The other comultiplications are obtained from $c$ as suitable perturbations.  In particular, if $f:X\rightarrow\Sigma \Omega X\wedge \Omega X$ is a map, then
$$c'=c+wf$$
is a comultiplication, where the sum is formed using the abelian group structure on $[X,X\vee X]$ which we have fixed with $c$. Of course, if we know already that $X$ admits other comultiplications, we may equally we start with them in place of $c$.
Finally, we come to understand such maps. We have 
$$[X,\Sigma\Omega X\wedge X]=[S^3\vee S^6,\Sigma\Omega X\wedge \Omega X]\cong \pi_3(\Sigma\Omega X\wedge \Omega X)\oplus\pi_6(\Sigma\Omega X\wedge \Omega X).$$
Using the Hilton-Milnor Theorem we have
$$\Omega X=\Omega\Sigma (S^2\vee S^5)\simeq \Omega S^3\times \Omega S^6\times \Omega S^8\times\dots$$
and with a little more work we get
$$\Sigma \Omega X\wedge \Omega X\simeq (\Sigma\Omega S^3\wedge \Omega S^3)\vee (\Sigma\Omega S^3\wedge \Omega S^5)\vee (\Sigma\Omega S^3\wedge \Omega S^5)\vee\dots$$
where the omitted terms are at least 8-connected.
Then $\Omega S^3$ is 2-conected and $\Omega S^5$ is 4-connected, so
$$\pi_3(\Sigma \Omega X\wedge \Omega X)=0.$$
On the other hand 
$$\pi_6(\Sigma \Omega X\wedge \Omega X)\cong \pi_6(\Sigma\Omega S^3\wedge \Omega S^3)\cong\pi_6(S^5\cup e^7\cup e^7\cup\dots).$$
A quick calculation in the cohomology on $\Omega S^3\wedge \Omega S^3$ shows that there is no Steenrod square $Sq^2:H^4(\Omega S^3\wedge \Omega S^3;\mathbb{Z}_2)\rightarrow H^6(\Omega S^3\wedge \Omega S^3;\mathbb{Z}_2)$, and since this operations detects the stable class $\eta$, we conclude that
$$\Sigma\Omega S^3\wedge \Omega S^3\simeq  (S^5\vee S^7\vee S^7)\cup\dots$$
(Note that $H^*\Omega S^3$ is a divided power algebra and is torsion free). In particular
$$\pi_6(\Sigma \Omega X\wedge \Omega X)\cong \pi_6(S^5\vee S^7\vee S^7)\cong \pi_6 S^5\cong \mathbb{Z}_2.$$
Putting everything together, comultiplications on $X=S^3\vee S^6$ are in bijective correspondence with the set
$$[X,\Sigma\Omega X\wedge \Omega X]\cong\pi_6S^5\cong\mathbb{Z}_2.$$
