(Hopf invariant, page 427 of A. Hatcher's Algebrac Topology):
Let $f: S^m\longrightarrow S^n$ with $m\geq n$. We can form a CW-complex $C_f$ by attaching a cell $e^{m+1}$ to $S^n$ via $f$. The homotopy type of $C_f$ depends only on the homotopy class of $f$. Thus for maps $f,g: S^m\longrightarrow S^n$, any invariant of homotopy type that distinguishes $C_f$ from $C_g$ will show that $f$ is not homotopic to $g$. In particular, when $m=n$, we can use $\deg f$, up to sign, as such an invariant since from the cellular chain complex of $C_f$, $ H_n(C_f;\mathbb{Z})=\mathbb{Z}/|\deg f|$.
Let $m=2n-1$. Then \begin{eqnarray*} &H^n(C_f;\mathbb{Z})=\text{Span}_\mathbb{Z}\{\alpha\}\cong\mathbb{Z},& \\ &H^{2n}(C_f;\mathbb{Z})=\text{Span}_\mathbb{Z}\{\beta\}\cong\mathbb{Z},&\\ &H^i(C_f;\mathbb{Z})=0, \text{ \ for } i\neq 0,n,2n,&\\ &H^*(C_f;\mathbb{Z})=\text{Algebra}_\mathbb{Z}\{1,\alpha,\beta\mid \alpha\smile \alpha=H(f)\beta\text{ for some }H(f)\in\mathbb{Z}\}.& \end{eqnarray*} Here we use the generalized definition of algebra over a commutative ring rather than merely over a field. The map $f$ induces a map $\tilde{f}: (D^{2n},\partial D^{2n}=S^{2n-1})\longrightarrow (C_f, S^n)$. This induces a homomorphism $\tilde f^*: H^{2n}(C_f, S^n;\mathbb{Z})\longrightarrow H^{2n}(D^{2n},S^{2n-1};\mathbb{Z})$. If we fix a generator $\sigma$ of $H^{2n}(D^{2n}, S^{2n-1};\mathbb{Z})$, then we can choose $\beta$ such that $\tilde f^*(\beta)=\lambda\sigma$ for a certain integer $\lambda\geq 0$.
Why $\lambda$ is nonzero? I want to prove $\tilde f^*$ is nontrivial but do not know how? Thanks.