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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.


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In your definition of $C_f$, shouldn't the sphere be $n$ dimensional, not $2n$ dimensional? – Jason DeVito Dec 19 '10 at 5:03
Youre right, thanks Jason. – Eric O. Korman Dec 19 '10 at 14:46
up vote 3 down vote accepted

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 $\pi_n(S^n)\rightarrow \pi_{n+1}(S^{n+1})$.


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,

$$ H(fg)\cdot \beta_{Cfg} = \alpha_{Cfg}^2=G^*(\alpha_{Cf}^2)=G^*(H(f)\cdot \beta_{Cf})=H(f)\cdot (\deg(g)\cdot \beta_{Cfg}).$$

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(Sorry this might be a little off, I'm kind of rushed but I saw nobody had answered so I wanted to say something) – Aaron Mazel-Gee Dec 19 '10 at 20:55
Thanks for the response. It may take me some time to digest it...from what you said does it follow that $G^*$ takes a generator $\beta$ of cohomology in degree $2n$ to $\deg g \beta$? – Eric O. Korman Dec 19 '10 at 22:35
Yes, it's because you can talk about the degree of a map of spheres either as the induced map on $Z$-(co)homology or as an element of $\pi_n(S^n)=Z$. Let me know if anything else doesn't make sense, I'm realizing this was pretty terse. – Aaron Mazel-Gee Dec 20 '10 at 4:08
Awesome, thanks! – Eric O. Korman Dec 20 '10 at 13:03

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.

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