 Solution

Polar deformation reduces maps from the circle into $G{L}_N$ to maps into $U(N)$. Define their integer invariant by the winding number of the determinant. It is unchanged under homotopy and stabilization, and determinants multiply under block sum, so it gives a homomorphism $K^1(S^1)\to \mathbb{Z}$. Every integer occurs, via $z\mapsto \mathrm{diag}(z^m,1,\dots ,1)$.

To prove injectivity, first consider a map with determinant winding zero. Its determinant has a continuous logarithm $ih(z)$ on the circle. Multiplying the first row by $e^{-ith(z)}$ gives a homotopy into $SU(N)$ at $t=1$. If $N=1$ that endpoint is already the identity. For $N\ge 2$, $SU(N)$ is simply connected: $SU(2)\cong S^3$, and the last-column fibrationwith locally trivializations from orthonormal completion, together with its homotopy exact sequence, proves this inductively because the sphere has zero first and second homotopy groups. Thus the remaining circle map is null-homotopic. For a map of winding $m$, multiply by $\mathrm{diag}(z^{-m},1,\dots ,1)$ and apply the zero-winding argument. This provesFor higher spheres, stable unitary homotopy gives $K^1(S^n)={\pi }_n(U)$: a map can be normalized by multiplying by the inverse of its value at the basepoint, a free homotopy because the unitary group is path connected. A free homotopy between based maps can be made based by replacing $H(x,t)$ with $H(x_0,t{)}^{-1}H(x,t)$, without altering its endpoints. The Bott periodicity theorem, in its unitary form ${\pi }_{j+2}(U)\cong {\pi }_j(U)$ with ${\pi }_0(U)=0$ and ${\pi }_1(U)=\mathbb{Z}$, givesThe circle case was proved directly above; the higher-sphere values use the stated general periodicity theorem.