Hopf invariant depends only on the homotopy type This is a question about Hopf invariant, discussed in Section 4.B of Hatcher's Algebraic Topology. http://pi.math.cornell.edu/~hatcher/AT/AT.pdf
Hatcher says that (in p.427 of the book) that the Hopf invariant of $f:S^{2n-1}\to S^n$ ($n>1$) depends only on the homotopy class of $f$, because under the homotopy equivalence $C_f\simeq C_g$ the chosen generators $\beta$ for $H^{2n}(C_f)$ and $H^{2n}(C_g)$ correspond, but I can't see why. Am I missing something? Thanks in advance.
 A: The generator $\beta$ is defined by the property that it is sent to a fixed generator of $H^{2n}(D^{2n},\partial D^{2n})$ by $H^{2n}(C_f,S^n) \to H^{2n}(D^{2n},\partial D^{2n})$, where the latter is induced by a map $(D^{2n},\partial D^{2n})\to (C_f, S^n)$ which is simply the attaching map.
But if $f\simeq g$, then you can construct a homotopy equivalence $C_f\to C_g$ which is compatible with these attaching maps, that is, such that the following square commutes up to homotopy
$$\require{AMScd}\begin{CD}(D^{2n},\partial D^{2n}) @>>> (C_f,S^n) \\
@VidVV @VVV\\
(D^{2n},\partial D^{2n}) @>>> (C_g,S^n)\end{CD}$$
that's essentially the content of proposition 0.18 (which Hatcher mentions in the paragraph you're referring to), or rather it follows from the proof of that proposition.
Since this square commutes up to homotopy, the corresponding square in cohomology commutes on the nose:
$$\begin{CD}H^{2n}(D^{2n},\partial D^{2n}) @<<< H^{2n}(C_f,S^n) \\
@AidAA @AAA\\
H^{2n}(D^{2n},\partial D^{2n}) @<<< H^{2n}(C_g,S^n)\end{CD}$$
which means that the generator for $C_g$ which is sent to the chosen generator of $H^{2n}(D^{2n},\partial D^{2n})$ is sent to the corresponding one for $C_f$, which is the claim.
