 Solution

Set $X={\mathbb{CP}}^4/{\mathbb{CP}}^1$ and $A={\mathbb{CP}}^2/{\mathbb{CP}}^1\cong S^4$. Let $u\in H^4(A;\mathbb{Z})$ be a generator and let $a\in H^4(X;\mathbb{Z})$ be the class restricting to $u$. The ring computation in part (a) gives

If $r\circ \iota \simeq {\mathrm{id}}_A$, then homotopy invariance of cohomology gives ${\iota }^{\ast }{r}^{\ast }(u)=u$, hence $r^{\ast }(u)=a$. But $u^2=0$ because $H^8(S^4;\mathbb{Z})=0$, whereas naturality of the cup product would givea contradiction. No such map $r$ exists.