Hopf invariant formula I'm having some trouble proving the following statement: let $g:S^{2n-1} \to S^{2n-1}, f: S^{2n-1} \to S^n$.  Then $H(f\circ g) = \deg g H(f)$ where $H(f)$ is the Hopf invariant.  The definition I am using for Hopf invariant is as follows:  let $C_f = D^{2n} \sqcup_f S^{n}$ where $D^{2n}$ is the $2n$ disk which we attach to $S^{n}$ via $f$.  Let $\alpha, \beta$ be generators for $H^n(C_f)$ and $H^{2n}(C_f)$ respectively.  Then $H(f)$ is defined by $\alpha \cup \alpha = H(f) \beta$.  I can prove the statement using the integral formula for $H(f)$ but would like to prove it cohomologically.
I think I should consider the map $G: C_{f\circ g} \to C_f$ that takes $x\in \partial D^{2n}$ to $g(x)$ and acts as the identity on everything else.  But then I don't know how to compute what $G^* \alpha$ is.
Thanks!
 A: So, if for $x\in \partial D^{2n}$ you take $G(x)=g(x)$, then probably you want to do "the same" map over the whole cone.  It follows from the definition of the union topology (or whatever it's called) that this extends to a continuous map.  Then, look at the induced map on the pairs $(K,K-D^{2n})$ (for either complex $K$), and here the map is just (up to confusing maps on pairs with maps on their quotients) the suspension of $g$.  Note that for both complexes, the inclusion $(K,\emptyset)\rightarrow (K,K-D^{2n})$ is a cohomology isomorphism in degree $2n$, and use the fact that suspension induces an isomorphism $\pi_n(S^n)\rightarrow \pi_{n+1}(S^{n+1})$.
Edit
A diagram might help:
    (Cfg,o)    --->     (Cf,o)
       |                  |
       |                  |
       V                  V
(Cfg,Cfg-D^2n) ---> (Cf, Cf-D^2n)

Here, the o's stand for $\emptyset$.  The vertical maps are inclusions of pairs, and the horizontal lines are really all just $G$ (as I defined it).  The top arrow induces $\alpha_{Cf} \mapsto \alpha_{Cfg}$, and the vertical maps are cohomology isomorphisms in degree $2n$.  On the bottom row, since $H^*(X,A)=\tilde{H}^*(X/A)$, this is basically a map $S^{2n}\rightarrow S^{2n}$, which by construction is exactly the suspension $Sg$ of $g$.  Then, $\deg(Sg)=\deg(g)$; that is, $S:\pi_{2n-1}(S^{2n-1})\rightarrow \pi_{2n}(S^{2n})$ is an isomorphism.  So this induces $\beta_{Cf}\mapsto \deg(g)\cdot \beta_{Cfg}$.  By naturality,
$$ H(fg)\cdot \beta_{Cfg} = \alpha_{Cfg}^2=G^*(\alpha_{Cf}^2)=G^*(H(f)\cdot \beta_{Cf})=H(f)\cdot (\deg(g)\cdot \beta_{Cfg}).$$
A: Another method would be to notice that the $n$-th and $2n$-th cohomology groups of $C_f$ and $C_{fg}$ are isomorphic. You can show that the isomorphism between the cohomology groups is then given via multiplication by $deg \hspace{0.5mm} g$ via the cellular boundary formula. 
