Folded product attachment has Hopf invariant two and null suspension

Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ be the positive generators from the two projections. By [F2], $a^2=b^2=0$ and $ab$ is the positive top generator. The map $\mathrm{\nabla }:W\to S^2$ extends over the cone on $\beta$ by the definition $\beta =\mathrm{\nabla }\alpha$; hence it extends to a map $q:S^2\times S^2=W{\cup }_{\alpha }{D}^4\to C_{\beta }$. On $W$, $q$ is the fold into the bottom $S^2$ of $C_{\beta }$, so $q^{\ast }x=a+b$. On the open $4$-cell, the extension is the characteristic disk of $C_{\beta }$ with degree $+1$ by [A1], so $q^{\ast }y=ab$.

Let $i:W\hookrightarrow S^2\times S^2$ be the $2$-skeleton inclusion. The characteristic disk extends $i\alpha$ over $D^4$, so $\mathrm{\Sigma }i\circ \mathrm{\Sigma }\alpha$ is null. Define $r=(\mathrm{\Sigma }{p}_1\vee \mathrm{\Sigma }{p}_2)\circ \mathrm{pinch}:\mathrm{\Sigma }(S^2\times S^2)⟶\mathrm{\Sigma }{S}^2\vee \mathrm{\Sigma }{S}^2,$ where $p_1,p_2$ are the projections and the pinch is [F3]. On the first suspended wedge summand, the first projection branch is the identity up to linear rescaling of the suspension coordinate and the second branch is constant; on the second summand the roles reverse. Thus $r\circ \mathrm{\Sigma }i\simeq {\mathrm{id}}_{\mathrm{\Sigma }W}$. Applying $r$ to the nullhomotopy in the first sentence gives $\mathrm{\Sigma }\alpha \simeq \ast$. Composing with $\mathrm{\Sigma }\mathrm{\nabla }$ gives $\mathrm{\Sigma }\beta \simeq \ast$.

In the cone $C_{\beta }$, $x^2=H(\beta )y$ by [F1]. Pulling back with step 1.1 yields $(a+b{)}^2=H(\beta )ab$. Since $a^2=b^2=0$ and degree-two classes commute, the left side is $2ab$. The class $ab$ is a generator, so $H(\beta )=2$.

For based maps $a,b:S^3\to S^2$, the group operation on ${\pi }_3(S^2)$ is represented by $(a\vee b)\circ p_{S^2}$, where $p_{S^2}:\mathrm{\Sigma }{S}^2\to \mathrm{\Sigma }{S}^2\vee \mathrm{\Sigma }{S}^2$ is the pinch in [F3]; the operation on ${\pi }_4(S^3)$ is represented by $p_{S^3}:\mathrm{\Sigma }{S}^3\to \mathrm{\Sigma }{S}^3\vee \mathrm{\Sigma }{S}^3$. In the quotient model ${\mathrm{\Sigma }}^{2}Z=(Z\times [0,1{]}^2)\mathrm{/}\text{\hspace{0.17em}⁣}\sim$, the map $\mathrm{\Sigma }{p}_Z$ pinches one suspension coordinate and $p_{\mathrm{\Sigma }Z}$ pinches the other. Write the square of suspension coordinates as a disk with its boundary collapsed. For each angle $\theta$, split the disk by the diameter at angle $\theta$, map the two half-disks to the two wedge summands using their boundary-collapsing rescalings to a full disk, and keep the first and second summands in that order. Choose these rescalings by rotating fixed rescalings of the two half-disks, so they vary continuously with $\theta$. As the diameter turns through a right angle, this gives a based homotopy between the two coordinate pinches; smashing this homotopy with $Z$ proves $\mathrm{\Sigma }{p}_Z\simeq p_{\mathrm{\Sigma }Z}$. Taking $Z=S^2$ and using $\mathrm{\Sigma }{S}^2\cong S^3$ gives $\mathrm{\Sigma }{p}_{S^2}\simeq p_{S^3}$. Since suspension preserves composition and sends a wedge of maps to the wedge of their suspensions, this compatibility gives $[\mathrm{\Sigma }(a+b)]=[\mathrm{\Sigma }a]+[\mathrm{\Sigma }b]$; hence $E:[a]\mapsto [\mathrm{\Sigma }a]$ is a homomorphism on ${\pi }_3(S^2)$. Now [F1, step 2.1] gives $[\beta ]=2[h]$, so $[\mathrm{\Sigma }\beta ]=E([\beta ])=2E([h])=2[\mathrm{\Sigma }h]$. Since $\mathrm{\Sigma }\beta$ is null by step 1.2, $2[{\eta }_3]=0$ for ${\eta }_3=\mathrm{\Sigma }h$.