 Solution

Put $m=2n$ and let $\alpha$ be the identity class of $S^m$. With the product-cell orientation used in part (i), form the folding map $\nabla :S^m\vee S^m\to S^m$, which is identity on each summand. The top cell of $S^m\times S^m$ is attached by the product of the two inclusions, so folding gives the attaching map $[\alpha ,\alpha {]}_W$. Thus there is a cellular mapwhich is the fold on the lower skeleton and degree one on the top relative cell.

Let $a,b\in H^m(S^m\times S^m;\mathbb{Z})$ be the classes from the two factors, and let $x\in H^m(C;\mathbb{Z})$ restrict to the generator of the bottom sphere. Orient the top class $z\in H^{2m}(C;\mathbb{Z})$ so that $F^{\ast }z=ab$. Then $F^{\ast }x=a+b$. Since $a^2=b^2=0$ and $m$ is even, graded commutative algebra givesThe map $F^{\ast }$ is an isomorphism in degree $2m$, because of its degree-one top-cell map. Therefore $x^2=2z$. By the definition of the Hopf invariant this provesThe shifted Lie convention in part (i) multiplies the geometric bracket by $(-1{)}^m=1$, so it gives the same square and the same invariant. Reversing the orientation chosen for the attached top cell would reverse both its generator and the numerical sign convention; with the standard orientation the value is the stated positive two.