Prove that if \(f,g:X\to S^{n-1}\) are continuous and both not surjective then \(f\) is homotopic to \(g\).
Solution
For the intended statement assume \(n\ge2\).
1. Every non-surjective continuous map\[ h:X\to S^{n-1} \]is null-homotopic.
Because \(h\) is not surjective, choose \[ p\in S^{n-1}\setminus h(X). \] Then \(h\) factors through the punctured sphere \[ S^{n-1}\setminus\{p\}. \] Stereographic projection from \(p\) gives a homeomorphism \[ \sigma:S^{n-1}\setminus\{p\}\longrightarrow\mathbb R^{n-1}. \] Choose a point \(x_0\in\mathbb R^{n-1}\) and define \[ H:X\times I\to S^{n-1}\setminus\{p\} \] by \[ H(x,t) =\sigma^{-1}\bigl((1-t)\sigma(h(x))+t x_0\bigr). \] This is continuous, with \[ H(x,0)=h(x) \] and \[ H(x,1)=\sigma^{-1}(x_0), \] independent of \(x\). Thus \(h\) is homotopic to a constant map.
2. The maps \(f\) and \(g\) are homotopic to constant maps\[ c_a(x)=a, \qquad c_b(x)=b \]for some \(a,b\in S^{n-1}\).
Apply 1 separately to the two non-surjective maps \(f\) and \(g\).
3. For \(n\ge2\), any two constant maps \(X\to S^{n-1}\) are homotopic.
Since \[ n-1\ge1, \] the sphere \(S^{n-1}\) is path-connected. Choose a path \[ \gamma:I\to S^{n-1} \] from \(a\) to \(b\). Then \[ K:X\times I\to S^{n-1}, \qquad K(x,t)=\gamma(t), \] is a homotopy from \(c_a\) to \(c_b\).
4. Hence for \(n\ge2\),\[ f\simeq g. \]
By 2, \[ f\simeq c_a \qquad\text{and}\qquad g\simeq c_b. \] By 3, \[ c_a\simeq c_b. \] Symmetry and transitivity of homotopy give \[ f\simeq c_a\simeq c_b\simeq g. \]
5. The source statement is false when \(n=1\).
Let \(X=\{*\}\) and identify \[ S^0=\{-1,1\}. \] Define \[ f(*)=1, \qquad g(*)=-1. \] Both maps are non-surjective. If they were homotopic, a homotopy \[ X\times I\to S^0 \] would give a path in \(S^0\) from \(1\) to \(-1\). No such path exists because \(S^0\) is discrete. Thus the unqualified statement fails for \(n=1\).