Show that a map \(f: X \to Y\) of connected CW complexes is a homotopy equivalence if it induces an isomorphism on \(\pi_1\) and its homotopy fiber \(F_f\) has \(\tilde{H}_*(F_f; \mathbb{Z}) = 0\).
Solution
Replace \(f:X\to Y\) by its mapping-path fibration \[ F_f\longrightarrow E_f\xrightarrow{p}Y, \] where \(E_f\simeq X\). The hypothesis on \(\pi_1\) says \[ \pi_1(E_f)\xrightarrow{\cong}\pi_1(Y). \] The long exact sequence therefore shows that \[ \pi_1(F_f)\to\pi_1(E_f) \] is the zero homomorphism.
By Exercise 10, the usual action of \(\pi_1(F_f)\) on every \(\pi_n(F_f)\), including conjugation on \(\pi_1(F_f)\), factors through \(\pi_1(E_f)\). Hence this action is trivial. In particular \(\pi_1(F_f)\) is abelian. Since \[ \widetilde H_1(F_f;\mathbb Z)=0, \] its abelianization is zero, so \[ \pi_1(F_f)=0. \]
Now \(F_f\) is simply connected and acyclic. If \(\pi_n(F_f)\ne0\) for a least \(n\ge2\), Hurewicz would give \[ \pi_n(F_f)\cong H_n(F_f)=0, \] a contradiction. Thus all homotopy groups of \(F_f\) vanish. The homotopy fiber has CW type, so Whitehead gives \[ F_f\simeq *. \] The fibration long exact sequence now shows that \(f\) induces isomorphisms on every homotopy group. Since \(X,Y\) are connected CW complexes, \[ \boxed{f\text{ is a homotopy equivalence}.} \]