Show that a map between simply-connected CW complexes is a homotopy equivalence if its mapping cone is contractible. Use the preceding exercise to give an example where this fails in the nonsimply-connected case.
Solution
Let \(f:X\to Y\) be a map of simply-connected CW complexes and suppose its mapping cone \(C_f\) is contractible. The cofibration sequence \[ X\xrightarrow f Y\longrightarrow C_f \] gives the long exact sequence in reduced homology. Since \(\widetilde H_*(C_f)=0\), the map \[ f_*:H_i(X;\mathbb Z)\xrightarrow{\cong}H_i(Y;\mathbb Z) \] is an isomorphism for every \(i\). The homology Whitehead theorem for simply-connected CW complexes then implies that \(f\) is a homotopy equivalence.
The simply-connected hypothesis is essential. Let \(A\) be any acyclic, noncontractible CW complex (for example an acyclic \(K(G,1)\) for a nontrivial acyclic group \(G\)), and take the constant map \[ f:A\longrightarrow *. \] Its mapping cone is \[ C_f=CA/A\cong\Sigma A. \] By the preceding exercise the suspension of an acyclic CW complex is contractible. Thus \(C_f\) is contractible, while \(f\) is not a homotopy equivalence because \(A\) is not contractible. This gives the required nonsimply-connected counterexample.