Use cohomology operations to show that the spaces \((S^1 \times \mathbb{CP}^\infty) / (S^1 \times \{x_0\})\) and \(S^3 \times \mathbb{CP}^\infty\) are not homotopy equivalent.
Solution
Put \[ X=(S^1\times\mathbb{CP}^\infty)/(S^1\times\{x_0\}), \qquad Y=S^3\times\mathbb{CP}^\infty. \] With \(\mathbb Z_2\)-coefficients, the relative Künneth theorem gives \[ \widetilde H^*(X;\mathbb Z_2) \cong H^*(S^1;\mathbb Z_2)\otimes \widetilde H^*(\mathbb{CP}^\infty;\mathbb Z_2). \] Let \[ a\in H^1(S^1;\mathbb Z_2), \qquad x\in H^2(\mathbb{CP}^\infty;\mathbb Z_2) \] be the standard generators. Then the unique nonzero class in \(H^3(X;\mathbb Z_2)\) is \[ u=a\times x. \] Using the Cartan formula, \[ Sq^2(u) =Sq^2(a\times x) =a\times Sq^2x, \] since \(Sq^1a=a^2=0\) and \(Sq^2a=0\) by instability. For the degree-two generator of \(\mathbb{CP}^\infty\), \[ Sq^2x=x^2. \] Hence \[ Sq^2(u)=a\times x^2\ne0 \in H^5(X;\mathbb Z_2). \tag{1} \]
On the other hand, the unique nonzero class in \(H^3(Y;\mathbb Z_2)\) is the pullback \[ v=\operatorname{pr}_1^*(\iota_3) \] of the fundamental class of \(S^3\). Naturality gives \[ Sq^2(v)=\operatorname{pr}_1^*(Sq^2\iota_3)=0, \tag{2} \] because \(H^5(S^3;\mathbb Z_2)=0\).
Any homotopy equivalence \(X\simeq Y\) would induce an isomorphism on mod-2 cohomology carrying the unique nonzero degree-three class to the unique nonzero degree-three class and commuting with \(Sq^2\). Equations (1) and (2) contradict this. Therefore \[ \boxed{X\not\simeq S^3\times\mathbb{CP}^\infty.} \]