- Lemma: \(S^n - \theset{p_i}_{i=1}^k = \bigvee_{k-1}S^{n-1}\), i.e. \(S^n\) minus \(k\) points is equal to \(k-1\) copies of of \(S^{n-1}\). Proof: \(S^n - \theset{p_1} \cong \RR^n\) by stereographic projection, so \(S^n - \theset{p_1, p_2 \cdots p_k} \cong \RR^n - \theset{p_2, \cdots p_k}\). WLOG, suppose none of these points are zero (otherwise, take a translation away from zero. This is affine and continuous.) Then fix 0 as the base point, and form \(k-1\) loops \(\alpha_i\), where the \(i\)th loop encircles \(p_i\). Then \(\RR^n\) deformation retracts onto \(\cup_{i=1}^{k-1} \alpha_i\), which is homeomorphic to \(\bigvee_{i=1}^{k-1} S^1\).
Solution
1. For \(k\ge1\), stereographic projection identifies \(S^n\) minus one chosen point with \(\mathbb R^n\), so removing \(k\) points from \(S^n\) reduces to removing \(k-1\) points from \(\mathbb R^n\).
Choose one of the removed points as the projection point.
2. If \(n\ge2\), \(\mathbb R^n\) minus \(r\) points deformation retracts onto a wedge of \(r\) copies of \(S^{n-1}\).
Choose disjoint small balls around the removed points and connect their boundary spheres to a common basepoint by disjoint arcs forming a tree. The complement deformation retracts onto the union of these boundary spheres and arcs; collapsing the tree gives the wedge of the \(r\) spheres.
3. Hence for \(n\ge2\),\[\boxed{S^n\setminus\{p_1,\dots,p_k\}\simeq\bigvee^{k-1}S^{n-1}.}\]
Apply <1>1–<1>2 with \(r=k-1\).