Let $R=\{(X,Y)\in \Bbb R^3\times \Bbb R^3: |X|=|Y|=1, \langle X,Y \rangle=0\}$.
Knowing that $R$ is not homeomorphic to $\Bbb S^2 \times \Bbb S^1$, prove that $R$'s projection onto first coordinate $Pr_1 : (X, Y ) \mapsto X $ doesn't have a continuous section, i.e. a continuous map $s:\Bbb S^2 \to R$ such that $Pr_1\circ s=Id_{\Bbb S^2}$.
Thank you for your help and comments.
What I tried so far:
I tried to use algebra to find a contradiction: If a continuous section exists, I tried to prove that $\Bbb S^1$ is an extension of $\Bbb S2$ through an exact sequence: $\Bbb S^1 \to R \to \Bbb S^2$ which will imply that $R \cong \Bbb S^2 \rtimes \Bbb S^1$. Hoping to go further and prove that the semi-direct product is direct and that we have a homeomorphism between $R$ and $\Bbb S^2 \times \Bbb S^1$
Well, let's suppose there is a section $s:S^2\rightarrow R$. Say $s$ has the form $S(x) = (f(x), g(x))$ for functions $f,g:S^2\rightarrow \mathbb{R}^3$. Being a section means $Pr_1\circ f = Id_{S^2}$, which, in turn, means that $f(x) = x$.
For any $x\in S^2$ and $\theta \in S^1$, we let $R_x(\theta)$ denote the rotation by angle $\theta$ counter-clockwise around the vector $x$. The map $S^2\times S^1\rightarrow SO(3)$ given by $(x,\theta)\mapsto R_x(\theta)$ is continuous (but not injective, since, for example, $(x,0)\mapsto I$ for any $x\in S^2$).
Note that if $(x,y)\in R$, then $(x,R_x(\theta) y)\in R$ because $R_x(\theta)$ is just acting as a rotation in the plane orthogonal to $x$ (where $y$ lives). Hence, we can construct a continuous map $\phi:S^2\times S^1\rightarrow R$ by $\phi(x,\theta) = (x, R_x(\theta)g(x))$.
I claim that $\phi$ is bijective. Momentarily believing this, because $S^2\times S^1$ is compact and $R$ is Hausdorff, this will imply $\phi$ is a homeomorphism, giving a contradiction.
So, why is $\phi$ injective? Well, assume $\phi(x_1, \theta_1) = \phi(x_2,\theta_2),$ that is, assume $(x_1, R_{x_1}(\theta_1)g(x_1)) = (x_2, R_{x_2}(\theta_2)g(x_2)$. It is then immediate that $x_1 = x_2$.
The equality coming from the second coordinate now looks like $R_{x_1}(\theta_1) g(x_1) = R_{x_1}(\theta_2)g(x_1)$. Multiplying both sides by the rotation matrix $R_{x_1}(-\theta_2)$, we see that $R_{x_1}(\theta_1 - \theta_2) g(x_1) = g(x_1)$. In other words, the rotation matrix $R_{x_1}(\theta_1 - \theta_2)$ has a real $1$-eigenvector $g(x_1)$ which is perpendicular to $x_1$ (which is another real eigenvector with eigenvalue $1$). It now follows that $R_{x_1}(\theta_1 - \theta_2) = Id$, so $\theta_1 = \theta_2$.
Now, why is $\phi$ surjective? Well, suppose $(x,y)\in R$. Let $\theta$ denote the angle between $y$ and $g(x)$. Then $\phi(x,\theta) = (x, R_x(\theta) g(x)) = (x,y)$.