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The manifold $\mathbb{CP}^2 \# -\mathbb{CP}^2$, the non-trivial $\mathbb{S}^2$ bundle over $\mathbb{S}^2$, is known to be diffeomorphic to the space that we will now describe. Represent $\mathbb{S}^3 \subseteq \mathbb{C}^2$ as pairs of complex numbers $(z_1, z_2)$ with $|z_1|^2 + |z_2|^2 = 1$. Let $\mathbb{S}^1$ act on $\mathbb{S}^3$ by $(w,(z_1, z_2)) \mapsto (wz_1, wz_2),$ where $w \in S^1$ is a complex number of modulus one. Let $S^1$ also act on $S^2$ by rotations around a fixed axis. Consider the space $M = \mathbb{S}^2 \times_{\mathbb{S}^1}\mathbb{S}^2$ obtained by taking the quotient of $\mathbb{S}^3 \times \mathbb{S}^2$ by the diagonal action of $S^1$.Then the manifold $M$ is diffeomorphic to $\mathbb{CP}^2 \# -\mathbb{CP}^2$.

Here I cannot find the homeomorphism between $M$ and $\mathbb{CP}^2 \# -\mathbb{CP}^2$. Please give me some ideas about this homeomorphism.

Thanks.

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    $\begingroup$ Cross-posted at math.stackexchange.com/questions/4834301/… (question closed). $\endgroup$
    – Mikhail Katz
    Commented Dec 31, 2023 at 11:43
  • $\begingroup$ I don't really get the question. As you've described it, M is the projectivization of the complex line bundle of Chern class 1 over $S^2$, and hence is exactly the non-trivial $S^2$ bundle over $S^2$. Are you asking why that's a connected sum of $\pm \mathbb{C}P^2$? $\endgroup$
    – Danny Ruberman
    Commented Dec 31, 2023 at 18:58
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    $\begingroup$ Maybe you'll like the explanation I gave here. $\endgroup$
    – mme
    Commented Dec 31, 2023 at 22:48

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