If a connected CW complex \(X\) retracts onto a subcomplex \(A\), show that \(\Sigma X \simeq \Sigma A \lor \Sigma(X/A)\).
Solution
Let \[ r:X\to A \] be a retraction and \[ q:X\to X/A \] the quotient map. Suspending and using the addition on maps out of a suspension gives \[ F=\Sigma r+\Sigma q:\Sigma X\longrightarrow \Sigma A\vee\Sigma(X/A). \] We show that \(F\) is a homology isomorphism.
Since \(r\circ i=\operatorname{id}_A\), the inclusion \[ i_*:\widetilde H_*(A)\to\widetilde H_*(X) \] is split injective, with splitting \(r_*\). The long exact sequence of the pair \((X,A)\) therefore breaks into split short exact sequences \[ 0\longrightarrow\widetilde H_j(A) \stackrel{i_*}{\longrightarrow} \widetilde H_j(X) \stackrel{q_*}{\longrightarrow} \widetilde H_j(X/A)\longrightarrow0. \] Hence \[ (r_*,q_*):\widetilde H_j(X)\xrightarrow{\cong} \widetilde H_j(A)\oplus\widetilde H_j(X/A). \tag{1} \] Under the suspension isomorphisms, \(F_*\) is exactly the map (1). Thus \(F\) induces an isomorphism on all reduced homology groups.
Both \(\Sigma X\) and \[ \Sigma A\vee\Sigma(X/A) \] are simply connected CW complexes because \(X\) is connected. The homology version of Whitehead’s theorem therefore gives \[ \boxed{\Sigma X\simeq\Sigma A\vee\Sigma(X/A).} \]