The homotopy fibre of an map $f \colon S^{2n-1} \to S^n$ I am trying to compute the cohomology of the homotopy fibre $F$ of a continuous map $f \colon S^{2n-1} \to S^n$ ($n$ even and non-zero Hopf invariant). It is easy to see that through the Serre spectral sequence the non zero groups are (with integer coefficients)
$$
H^{n-1}(F) = \mathbb Z,\quad H^k(F) = \mathbb Z_m 
$$
for $k = 2n-1 + j(n-1)$ and $j=0,1,2,\dots$ ($m=1$ means here that all that the latter groups are all zero). Someone mentioned to me that $m$ is just the Hopf-invariant of $f$. He didnt had a proof and I can not show or see this.
 A: In the following, any identity may actually be up to a sign.  I don't want to keep track of that.
We are thinking about the fibration sequence
$$ \Omega S^n \to F \to S^{2n-1} \xrightarrow{f} S^n.$$
It may be little easier to think about the homology Serre spectral sequence of $\Omega S^n\to F\to S^{2n-1}$ rather than the cohomology one: they are basically dual to each other.  We are interested in the first non-trivial differential:
$$ d\colon E^{2n-1,0}=H_{2n-1}S^{2n-1} \to E^{0,2n-2}= H_{2n-2}\Omega S^n.$$
A "standard" argument shows that $d$ sends the generator (which is the Hurewicz image of the identity map of $S^{2n-1}$) to the Hurewicz image of 
$$\tilde{f}\colon S^{2n-2} \to \Omega S^n,$$
the adjoint to $f$.
We have that $H_*\Omega S^n$ is a polynomial ring on one generator $a$ in (odd) degree $n-1$; the product is the Pontryagin product associated to the H-space structure on $\Omega S^n$ given by composition of loops.  Note that although this is a polynomial ring, it is not commutative in the graded sense: for instance, since $a$ is in odd degree, the graded commutator gives $[a,a]= a.a - (-1) a.a = 2a^2$.
So $H_{2n-2}\Omega S^n \approx \mathbb Z$, generated by $a^2$.  Given $f\colon S^{2n-1}\to S^n$, we obtain an invariant $H'(f)\in \mathbb Z$ by $h(\tilde f)= H'(f) a^2$, where $h\colon \pi_* \Omega S^n\to H_*\Omega S^n$ is the Hurewicz map.  I claim that $H'(f)$ is the same as the Hopf invariant of $f$ (up to a sign).
There's surely a slick proof of that, but I don't remember it.  Here's a sketch of a dirty proof: 


*

*Show that $H'$ is a group homomorphism (this basically amounts to the fact that the pinch map $S^{2n-1}\to S^{2n-1}\vee S^{2n-1}$ is a suspension).  (Remember that the Hopf invariant $H$ is also a homomorphism).

*By Serre, we know that $\pi_{2n-1}S^n\approx \mathbb Z\oplus \text{torsion}$.  Because $H'$ is a homomorphism, it factors through the torsion-free quotient.  Thus it suffices to show $H'(f)=H(f)$ for one element $f$ of infinite order.

*Let $f=[\iota,\iota]$, the Whitehead product of the identity map of $S^n$ with itself.  The adjoint $\tilde f$ is (up to sign) the "Samelson product" $\{ \tilde\iota, \tilde\iota \}$.  Here $\tilde\iota\colon S^{n-1}\to \Omega S^n$ is the adjoint of identity: the Samelson product on $\pi_p \Omega X\times \pi_q \Omega X\to \pi_{p+q}\Omega X$ is the bilinear map on homotopy groups induced by the commutator operation on the "group" $\Omega X$.
The Hurewicz map $h\colon \pi_*\Omega X\to H_*\Omega X$ takes Samelson products to commutators in the Pontryagin ring.
It is standard that $H(f)=2$.  On the other hand, $$h(\tilde f)= h(\{\tilde \iota, \tilde \iota\})= [h(\tilde \iota), h(\tilde\iota)]=[a,a]=2a^2,$$ so $H'(f)=2$.
