Relative homotopy of a disk boundary pair

The homotopy $H(x,t)=(1-t)x+t{x}_0$ stays in the disk by convexity, is continuous by F1 and fixes x0. Applying it to any based cubical representative proves every positive absolute group of $D^m$ trivial. Hence for k≥2 the exact segment $0\to {\pi }_k(D^m,S^{m-1})\stackrel{\mathrm{\partial }}{\to }{\pi }_{k-1}(S^{m-1})\to 0$ makes the boundary map an isomorphism.

For $2\le k<m$, apply F3 with source sphere dimension k−1 and target dimension m−1, using F5, to make the target of this isomorphism zero. For k=m, F4 identifies it with Z. Restricting the oriented characteristic map $\mathrm{i}\mathrm{d}:D^m\to D^m$ gives the oriented boundary identity, whose degree is +1. Thus its relative class is the asserted generator; for example when m=2 the boundary sends that disk to the once-traversed oriented circle, with coefficient 1.

For m≥2, F3 at source dimension zero makes $S^{m-1}$ path-connected. The low tail $0={\pi }_1(D^m)\to {\pi }_1(D^m,S^{m-1})\to {\pi }_0(S^{m-1})$ then makes the relative pointed set a singleton by F2. For m=1, write the disk as [-1,1] and x0=1 (reflection gives the other case). A representative α starts at a∈{−1,1} and ends at 1. The interpolation $(1-s)\alpha (t)+s((1-t)a+t)$ is continuous, stays in the interval, and fixes both endpoints. Thus all representatives with the same a are equivalent. A relative homotopy cannot change a, since a continuous path into the discrete two-point set is constant. Representatives $t\mapsto 1$ and $t\mapsto 2t-1$ therefore give exactly two classes.