Based and free homotopy classes of maps between spheres agree

A rotation in a plane containing $f(\ast )$ and the target basepoint takes $f(\ast )$ to that basepoint and is joined to the identity by varying its angle. If the two points coincide use the identity, and if antipodal choose any perpendicular unit vector, available since $k+1\ge 2$. Postcomposition gives a free homotopy from $f$ to a based map. This proves surjectivity.

For unit vectors $a,b$ with $a\cdot b>-1$, put $K=b{a}^T-a{b}^T$ and $Q(a,b)=I+K+K^2\mathrm{/}(1+a\cdot b)$. On the plane spanned by $a,b$, this is the rotation taking $a$ to $b$; on its orthogonal complement it is the identity. Direct multiplication gives $Q(a,b)a=b$ and $Q(a,b{)}^{T}Q(a,b)=I$; its determinant is one and $Q(a,a)=I$. The formula is continuous even at $a=b$. For a continuous path $p:I\to S^k$, choose a finite subdivision so that $p(t)\cdot p(t_j)>-1$ on each subinterval, using uniform continuity. Set $P_0=I$ and inductively $P_t=Q(p(t_j),p(t))P_{t_j}$. Then $P_t$ is continuous in $SO(k+1)$ and $P_{t}p(0)=p(t)$.

Suppose based $f_0,f_1$ are freely homotopic by $H$. Apply step 1.2 to $p(t)=H(\ast ,t)$ and define $\widehat{H}(x,t)=P_t^{-1}H(x,t)$. This is a based homotopy from $f_0$ to $P_1^{-1}{f}_1$. Since $p(0)=p(1)=\ast$, the matrix $P_1$ fixes the basepoint vector and restricts to an element of $SO(k)$ on its perpendicular subspace. Every element of $SO(k)$ is joined to the identity by plane rotations: successively rotate its first column to the first coordinate vector, then its second column within the perpendicular complement, continuing until the final one-dimensional block, which is $+1$ because the determinant is one. At each stage an antipodal column is handled by a rotation through $\pi$ in a two-plane; for $k=1$ the group is already the identity. Reversing the finite sequence and varying the angles gives the required path in the stabilizer of $\ast$. Postcomposing $f_1$ with that path joins $P_1^{-1}{f}_1$ to $f_1$ through based maps. Concatenation with $\widehat{H}$ proves injectivity.

Surjectivity and injectivity prove the assertion, including $n=k=1$, where the final stabilizer is trivial. All selections are finite. Via [F3] this is the stated bijection for the cubical group ${\pi }_n(S^k)$.