Let \(f: X \to S^n\) be a continuous map that is not surjective. Prove that \(f\) is nullhomotopic.
Solution
1. Since \(f\) is not surjective, there exists a point \(p \in S^n \setminus f(X)\).
definition of non-surjectivity.
2. Let \(s_0 = -p \in S^n\) be the antipodal point to \(p\).
definition of antipodal point on \(S^n\).
3. For all \(x \in X\) and all \(t \in [0, 1]\), \((1 - t)f(x) + t s_0 \neq 0\) in \(\mathbb{R}^{n+1}\).
3.1. Suppose \((1 - t)f(x) + t s_0 = 0\) for some \(t \in [0, 1]\) and \(x \in X\).
hypothesis for contradiction.
3.2. If \(t = 0\), then \(f(x) = 0\), impossible since \(f(x) \in S^n \implies \|f(x)\| = 1\).
norm on \(S^n\).
3.3. If \(t = 1\), then \(s_0 = 0\), impossible since \(\|s_0\| = 1\).
norm on \(S^n\).
3.4. For \(0 < t < 1\), \((1 - t)f(x) = -t s_0\).Taking norms gives \((1 - t) = t \implies t = 1/2\).
\(\|f(x)\| = \|s_0\| = 1\).
3.5. Then \(f(x) = -s_0 = -(-p) = p\).
\((1-t)f(x) = -t s_0\) with \(t = 1/2\).
3.6. But \(p \notin f(X)\), so \(f(x) \neq p\), a contradiction.
1.
1.7. Thus \((1 - t)f(x) + t s_0 \neq 0\) for all \(x \in X, t \in [0, 1]\).
2, 3, and 6.
4. Define \(H: X \times [0, 1] \to S^n\) by\[ H(x, t) = \frac{(1 - t)f(x) + t s_0}{\|(1 - t)f(x) + t s_0\|}. \]
4.1. \(H\) is well-defined because the denominator is non-zero by
3.
3.
3.2. \(H\) is continuous as a composition of continuous functions (vector addition, scalar multiplication, norm, and quotient).
continuity of linear operations and norm on \(\mathbb{R}^{n+1} \setminus \{0\}\).
3.3. For \(t = 0\): \(H(x, 0) = \frac{f(x)}{\|f(x)\|} = f(x)\) since \(\|f(x)\| = 1\).
\(f(x) \in S^n\).
3.4. For \(t = 1\): \(H(x, 1) = \frac{s_0}{\|s_0\|} = s_0\) since \(\|s_0\| = 1\).
\(s_0 \in S^n\).
3.5. Hence \(H\) is a homotopy between \(f\) and the constant map \(x \mapsto s_0\).
<2>1–<2>4.
5. Therefore \(f\) is nullhomotopic.
4.
6. Q.E.D.
5.