Show that any non-surjective map \(f: X \rightarrow S^n\) is homotopic to the constant map.
Solution
Goal: Let \(X\) be an arbitrary topological space, and let \(f \colon X \to S^n\) be a continuous map (\(n \ge 1\)). Prove that if \(f\) is not surjective, then \(f\) is nullhomotopic (homotopic to a constant map).
1. Pick a point omitted by \(f\).
1.1. Since \(f\) is not surjective, there exists a point \(p \in S^n\) such that \(p \notin f(X)\), so \(f(X) \subseteq S^n \setminus \{p\}\).
1.2. Without loss of generality, by rotating \(S^n\) if necessary, let \(p = -N = (0, \dots, 0, -1)\) (the south pole), and let \(N = (0, \dots, 0, 1)\) (the north pole).
1.3. Since \(f\) is not surjective, \(f(X) \neq S^n\), so some point \(p\) is omitted; a rotation of \(S^n\) carries \(p\) to the south pole without changing the homotopy class of \(f\).
2. Show that the punctured sphere \(S^n \setminus \{p\}\) is contractible.
2.1. Stereographic projection from the pole \(p\) gives a homeomorphism \(\phi \colon S^n \setminus \{p\} \to \mathbb{R}^n\).
2.2. Euclidean space \(\mathbb{R}^n\) is convex, hence contractible to the origin \(0 \in \mathbb{R}^n\) via the straight-line homotopy \(h(y, t) = (1-t)y\).
2.3. Pulling back via \(\phi\), define \(H_0 \colon (S^n \setminus \{p\}) \times [0, 1] \to S^n \setminus \{p\}\) by \(H_0(z, t) = \phi^{-1}((1-t)\phi(z))\).
2.4. \(H_0(z, 0) = z = \operatorname{id}_{S^n \setminus \{p\}}(z)\), and \(H_0(z, 1) = \phi^{-1}(0) = N\) (constant map to the antipodal pole \(N\)).
2.5. \(H_0\) is continuous because \(\phi\), \(\phi^{-1}\), and scalar multiplication are continuous.
2.6. The map \(H_0\) is a homotopy from the identity on \(S^n \setminus \{p\}\) to the constant map at \(N\), so \(S^n \setminus \{p\}\) is contractible to the point \(N\).
3. Construct the nullhomotopy for \(f\).
3.1. Define \(H \colon X \times [0, 1] \to S^n\) by \(H(x, t) = H_0(f(x), t)\).
3.2. Equivalently, using normalized straight-line geodesics in \(\mathbb{R}^{n+1}\): Since \(f(x) \neq -N\) for all \(x \in X\), \((1-t)f(x) + t N \neq 0\) for all \(t \in [0, 1]\), so \[H(x, t) = \frac{(1-t)f(x) + t N}{\|(1-t)f(x) + t N\|}\] is well-defined, continuous, and lands in \(S^n\).
3.3. Check endpoints:
\(H(x, 0) = \frac{f(x)}{\|f(x)\|} = f(x)\),
\(H(x, 1) = \frac{N}{\|N\|} = N = c_N(x)\). <2>4. Thus \(H\) is a homotopy between \(f\) and the constant map \(c_N \colon X \to S^n\).
3.5. The endpoint computations in 3 show \(H(\cdot, 0) = f\) and \(H(\cdot, 1) = c_N\), and \(H\) is continuous, so \(H\) is a homotopy from \(f\) to the constant map.
4. Q.E.D.
4.1. 3 proves that \(f\) is homotopic to the constant map \(c_N\).