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I have some questions about the space $\textbf{SO(3)}$ (special orthogonal transformations in 3-dimensions). I understand that this transformations represent rotations in 3D space fixing the origin and also I understand that $\textbf{SO(3)}$ is homeomorphic to $\textbf{RP(3)}$ (real-projective 3D-space). A proof for this homeomorphism can be found in: prove $RP^3\cong SO(3)$

Now imagine an orthogonal base at the origin ($e_1$, $e_2$, $e_3$). To specify a rotation you have to first choose where in the unit sphere you want to send $e_1$, once you choose that direction you have to choose where to send $e_2$ around a circle, and once you choose that you are done (because of the orthogonality). So with this analysis it seems that the group of rotations should be homeomorphic to $S^2 \times S$ (sphere cross the cirle). But also $\textbf{SO(3)}$ is homeomorphic to $\textbf{RP(3)}$.

My question is: Are $S^2 \times S$ and $RP^3$ really homeomorphic? and if not where is the mistake on my reasoning?, if they are, is there a more simple homeomorphism to show that (without the use of $\textbf{SO(3)}$ and more in the spirit of cutting and gluing)?

Thanks in advance. :)

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    $\begingroup$ I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration. $\endgroup$ – Daniel Schepler Dec 1 '18 at 0:11
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They are not even homotopy equivalent. We have $\pi_1(\textbf{RP(3)})= \mathbb{Z}_2$ (see Sammy Black's comment to An intuitive idea about fundamental group of $\mathbb{RP}^2$), but $\pi_1(S^2 \times S^1 )= \mathbb{Z}$.

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