I have seen two definitions of the Hopf invariant given:
(1) Cohomological Definition: Let $S^{n}$ denote the oriented $n$-sphere, where $n \geq 2$. Let there be given a map $f:S^{2n-1} \rightarrow S^{n}$. Consider $S^{2n-1}$ as the boundary of an oriented $2n$-cell, and form the cell complex $C=S^{n} \cup_{f} e^{2n}$. This is the complex formed from the disjoint union of $S^{n}$ and $e^{2n}$ by identifying each point in $S^{2n-}=\overset{\circ}{e^{2n}}$ with its image under $f$. Then the integral cohomology of $K$ is zero except for dimensions $0,n, 2n$, and $\mathbb{Z}$ is in those three dimensions. Denote by $\sigma$ and $\tau$ the generators determined by the given orientations of the cohomology groups in dimensions $n$ and $2n$ respectively. Then the cup product square $\sigma^{2}$ is some integral multiple of $\tau$.
The Hopf invariant of $f$ is the integer $h(f)$ such that $\sigma^{2}=h(f) \cdot \tau$. (See Mosher/Tangora pg. 33).
(2) $K$-theory: Here we use the same cell-complex formulation, but instead of considering integral cohomology we have $\pi: K \rightarrow S^{n}/K \cong S^{2n}$, which is the quotient map that collapses $S^{n}$. We then have a sequence $S^{2n-1} \overset{f}{\rightarrow} S^{n} \overset{i}{\rightarrow} C \overset{\pi}{\rightarrow} S^{2n} \overset{\Sigma f}{\rightarrow} S^{n+1}$.Specializing to $n$ even, we reduce to the $K$-theory of $(X, S^{n})$. We know that $\tilde{K}^{1}(S^{2n})=\tilde{K}^{1}(S^{n})=0$.Hence we have a short exact sequence $0 \rightarrow \tilde{K}^{1}(S^{2n}) \overset{\pi^{*}}{\rightarrow} \tilde{K}(C) \overset{i^{*}}{\rightarrow} \tilde{K}(S^{n}) \rightarrow 0$. We denote the generator of $\tilde{K}(S^{n})$ by $i_{n}$ and the generator of $\tilde{K}(S^{n})$ by $i_{2n}$. Select some $a \in \tilde{K}(C)$ such that $i^{*}(a)=i_{n}$ and let $b=\pi^{*}(i_{2n}) \in \tilde{K}(C)$. Our $K$-theory short exact sequence shows that $\tilde{K}(C)$ is free abelian with generators $a,b$.Now, since any square in $\tilde{K}(S^{n})$ vanishes we have $i^{2}_{n}=0$. Hence, the Hopf invariant is the integer $h(f)$ such that $a^{2}=h(f) \cdot b$.
I am trouble seeing why these two definitions are equivalent (at least up to sign). Could someone give an indication as to why this is true? Thanks very much.