The Hopf invariant is an isomorphism for $\pi_3 (S^2)$ Does any one have a reference to a proof that the Hopf invariant classifies the homotopy classes of maps from $\mathbb{S}^3$ to $\mathbb{S}^2$.
It is quite standard to find a proof that the Hopf invariant is a homotopy invariant. However, after searching MathSciNet for all the books covering the Hopf invariant, I could not find any proofs more recent than Hopf's original 1931 paper written in German. The best that I have found are some statements without any proof or reference.
 A: As Robert Bryant's comments shows, showing this isomorphism is a standard beginning grad student exercise.  More deeply, in the 1950's, Ioan James put the Hopf invariant into a more general context with his EHP long exact sequence.  (E for suspension - it helps if you are speaking German - and P for Pontryagin.) Lots of textbooks discuss this; George Whitehead's Elements of Homotopy (a Springer Grad Text in Math) does a particularly thorough job.  Joe Neisendorfer's recent book Algebraic Methods in Homotopy Theory shows how far one can go with these ideas.   You can get more references by Googling EHP sequence.
A: Theorem. The Hopf invariant is a non-zero group homomorphism $\mathcal H:\pi_{4n-1}(\mathbb S^{2n})\to\mathbb Z$ and it is an isomorphism only when $n=1$.
For a proof that $\mathcal H$ is a group homomorphism, 
see Proposition 4B.1 in [3].
Hopf, in [4] (see Satz II and Satz II') proved that for any $n$, there is a map
$h:\mathbb S^{4n-1}\to\mathbb S^{2n}$ with $\mathcal H h\neq 0$ and hence the homomorphism
$\mathcal H:\pi_{4n-1}(\mathbb S^{2n})\to\mathbb Z$ is non-zero.
Since the Hopf invariant of the Hopf fibration $h:\mathbb S^3\to\mathbb S^2$ equals $1$ (Example 17.23 in [2]),
$\mathcal H:\pi_3(\mathbb S^2)\to\mathbb Z$ is an isomorphism. However, for $n\geq 2$ the Hopf invariant is never an isomorphism.
Indeed, Adams [1] proved that mappings with Hopf invariant equal $1$ exist only when $n=1,2$ and $4$,
so these are the only cases when one may suspect $\mathcal H$ to be an isomorphism, but
$\pi_7(\mathbb S^4)=\mathbb Z\times\mathbb Z_{12}$ and  $\pi_{15}(\mathbb S^8)=\mathbb Z\times\mathbb Z_{120}$, so $\mathcal H$ cannot be an isomorphism.
[1] J. F. Adams, J.,
On the non-existence of elements of Hopf invariant one.
Ann. of Math. 72 (1960), 20-104.
[2] R. Bott, L. W. Tu, L. W.
Differential forms in algebraic topology.
Graduate Texts in Mathematics, 82. Springer-Verlag, New York-Berlin, 1982.
[3] A. Hatcher,
Algebraic topology.
Cambridge University Press, Cambridge, 2002.
[4] H. Hopf,
Über die Abbildungen von Sphären auf Sphären niedrigerer Dimensionen. 
Fundamenta Math. 25 (1935), 427-440.
A: It depends on your definition of Hopf invariant. One definition is to look at the cup product structure in the integral cohomology of the cofiber. It then takes a bit of work to show it is a homomorphism, but once done the fact that the Hopf invariant of the Hopf map is $1$ is the standard calculation for projective spaces. 
A: As Robert Bryant and Nicholas Kuhn have indicated, this is standard stuff that is in many textbooks.  One thing that might be tripping you up is that the original definition of the Hopf invariant is very geometric (involving linking numbers), while most modern treatments use alternative definitions.  If you're interested in doing things in a more geometric way, see my notes "Homotopy groups of spheres and low-dimensional topology", which are available on my webpage here.
A: You might like to see this result on $\pi_3(S^2)$ as a special case of a result on $\pi_3(S(K(G,1))$ for any group $G$, where $S$ is suspension. This exposition shows the involvement of a nonabelian tensor product $G \otimes G$ of a group $G$  and how to describe the Whitehead product $\pi_2 \times \pi_2 \to \pi_3$ for $S(K(G,1))$,  and composition with the Hopf map,  in these terms. 
