- Show that if \(X\mapsvia{f} X^n\) is not surjective, then \(f\) is nullhomotopic.
Solution
1. The notation \(X^n\) is not defined in the statement, so the target space is not determined.
2. For spheres, if \(f:S^n\to S^n\) is not surjective, then \(f\) is null-homotopic.
Choose \(p\in S^n\setminus f(S^n)\). Then \(f\) factors through \(S^n\setminus\{p\}\cong\mathbb R^n\), which is contractible. Hence the factorization, and therefore \(f\), is null-homotopic.