J-homomorphism

Remarks

The two clauses agree on $S^r \times S^{n-1}$, since $f(x)$ preserves the unit sphere, so $f(x) v \in \partial D^n$. The map $J$ is a well-defined group homomorphism.

Example: for $n = 2$ and $r = 1$, let $f \colon S^1 \to SO(2)$ send $z$ to rotation by $z$, viewing $\R^2$ as $\C$. Then $J(f) \colon S^3 \to S^2$ is the Hopf map, up to sign and homotopy, and it generates $\pi_3(S^2) \cong \Z$.

Non-example: suppose $f(x)$ is allowed to be any invertible linear map. Then $f(x) v$ need not lie on $\partial D^n$ for $v \in S^{n-1}$, and may even leave $D^n$, so the two clauses do not define a map.

The maps $J$ commute with the stabilisation maps $SO(n) \to SO(n+1)$ and with suspension, so they give the stable J-homomorphism $J \colon \pi_r(SO) \to \pi_r^s$.