I'm having some trouble with a problem about the Hopf construction, in the exercises for Ch. 4 of Mosher & Tangora. Given a map $g : S^{n-1} \times S^{n-1} \rightarrow S^{n-1}$, we get a map $h(g) : S^{2n-1} \rightarrow S^n$ by considering

$S^{2n-1} = S^{n-1} * S^{n-1} = S^{n-1} \times I \times S^{n-1} / \sim$

$S^n = S(S^{n-1}) = S^{n-1} \times I / \sim$

and by putting $h(g)(a,t,b) = (g(a,b),t)$. There's an easy homotopy invariant $(a,b)$ of the map $g$, the degree of the map when restricted to either factor (times any point of the other $S^{n-1}$). The problem asks me to show that the Hopf invariant of the map $h(g)$ is $H(h(g))=ab$. This is defined by $s^2=H(h(g))t$ for generators $s\in H^n(X)$ and $t\in H^{2n}(X)$, where $X=S^n \cup_{h(g)} S^{2n}$ (here $h(g)$ is the attaching map).

I'm trying to mimic a proof in the chapter, which constructs a map with Hopf invariant 2 for even $n$ which is really just the Whitehead square of the identity map. They use the following diagram ($f$ is the folding map, $F$ is induced from $f$, $g$ is the attaching map to get from $S^n \vee S^n$ to $S^n \times S^n$, the vertical maps are inclusions, and $K=S^n \cup_{fg} e^{2n}$ is the complex in which we need to calculate cup products):

```
g f
S^{2n-1} ----> S^n V S^n ------------> S^n
| |
| |
| | i
| |
V F V
S^n x S^n = ( S^n V S^n ) U_g e^{2n} -----> K
```

[**Edit**: Uh oh, how do I get it to put stuff in a uniformly spaced font? I'm too illiterate to understand the "formatting reference" -- any help??]

Note that $F$ exists because the composition $ifg:S^{2n-1} \rightarrow K$ is nullhomotopic. We know that $f^*s=s_1 + s_2$, where $s$ and $s_i$ represent the obvious cohomology generators. Denoting 2n-dim. cohomology generators of $K$ and $S^n \times S^n$ by $t$ and $r$ respectively, since $F$ induces an isomorphism in $H^{2n}$, (we can choose orientations so that) $F^* t=r$. And by the Kunneth formula, in $H^*(S^n \times S^n)$ we have $s_1 s_2=r$ and $s_1^2=s_2^2=0$. So now we calculate: $F^*(s^2) = (F^*s)^2 = (f^*s)^2 = (s_1 + s_2)^2 = s_1^2 + s_1 s_2 + s_2 s_1 + s_2^2 = 2r$ (assuming $n$ is even; otherwise $F^*(s^2)=0$).

So somehow they're using things they know about the cohomology of $S^n \times S^n$ to compute cup products in $H^*(K)$. This seems like it's possible because the attaching map $fg$ factors through $S^n \vee S^n$. In the problem I'm stuck on, I can't figure out what should be the analogous factorization. Alternatively, I was thinking that I could possibly use Poincare duality and just try to find the self-intersection of the copy of $S^n$ inside of $K$, but that seem too silly/gimmicky. Another fact I know is that $A * B \simeq \Sigma(A \wedge B)$. But suspension doesn't preserve cup products, so I don't think this could help either...

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