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This is the exercise (6.14) of Greenberg-Harper Algebraic Topology book.

I already showed continuity but I have problems for shows injectivity and surjectivity. For injectivity, if I assume $R_\theta \circ A(x)=R_{\theta^{\prime}} \circ A(x^\prime)$ (with $-\pi\leq \theta <\pi$) then $R_\theta \cdot A(x)=R_{\theta^{\prime}} \cdot A(x^\prime)$ (as matrix). But I dont Know how to conclude that $\theta=\theta^{\prime}$ and $x=x^\prime$.

Other form to show its, is defining the inverse of map. I think to define $\psi: SO(3)\to S^{1}\times S^{2}$ by $T \mapsto (T\cdot (A(T(e_1))^{-1}, T(e_1))$ (Here $T(e_1)$ denotes first column of $T$) and see if $T\cdot (A(T(e_1))^{-1}$ is the form $$T\cdot (A(T(e_1))^{-1}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta &-\sin \theta\\ 0& \sin \theta &\cos \theta \end{bmatrix}$$ with finality to identify $T\cdot (A(T(e_1))^{-1}$ with an angle. But, I dont know if a matrix of $T\cdot (A(T(e_1))^{-1}$ has this form. I tried to prove its. Is easy see that $A((T(e_1))^{-1} \cdot T$ has a form described above. But what I want is that $T\cdot (A(T(e_1))^{-1}$ but to have the form described above.

I also tried to do the same by taking $T(e_1)$ as the first row instead of the first column but still can't see that it has the required form.

Danimenru
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  • Isn't it Greenberg and Harper? What happened to Harper? –  May 15 '21 at 04:45
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    Yes, Greenberg and Harper. – Danimenru May 15 '21 at 04:47
  • Suppose the hairy ball theorem is false, then you can construct a global orthonormal frame on the tangent space of the sphere, and then $\theta$ determines the second vector of a positive orthonormal basis (second column of your rotation matrix, the first would be $x$ itself). Finally, you arrive at a homeomorphism that should not exist by a previous example/exercise. I think In here you can arrive at essentially the same bijection in a more geometric-friendly way. – Gustavo Labegalini May 15 '21 at 05:41

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Under the ordered basis $(x, t(x), n(x))$, the rotation is given by $$ \Theta = \begin{pmatrix}1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos \theta \end{pmatrix},$$

thus in the standard basis, $$ R_{\theta} = A \Theta A^{-1}$$ and thus \begin{align} \phi (\theta, x) &= R_\theta A(x) \\ &= A(x) \Theta \\ &= \begin{pmatrix}x & \cos\theta t(x) + \sin\theta n(x) & -\sin\theta t(x) + \cos\theta n(x) \end{pmatrix}. \end{align}

If $\phi(\theta, x) = \phi (\theta', x')$, by checking the first row we have $x = x'$. Since $\phi(\theta, x) = R_\theta A(x)$, we have $R_\theta = R_{\theta'}$ and thus $\theta = \theta'$. Hence $\phi$ is injective.

Surjectivity follows from the fact that any $T\in SO(3)$ is a rotation about an vector $x\in \mathbb S^2$. Then $T = R_\theta A(x)$ for some $\theta, x$ (see here for a proof).

Arctic Char
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    Very thanks. I have a questao. Why $R_\theta A(x)=(x ~~~~ \cos\theta t(x)+\sin \theta n(x) −\sin\theta t(x)+\cos\theta n(x) )$? I understand from statement of exercise that $R_\theta$ is a rotation about axis $x$ and hence $$R_\theta=\begin{bmatrix} 1 & 0 & 0\ 0 &\cos \theta & -\sin \theta \ 0 & \sin \theta &\cos \theta \end{bmatrix}$$ But $R_\theta A(x) $not result equal to the vector above . Or am I misunderstanding it? – Danimenru May 15 '21 at 04:51
  • I am interpreting $A (x) = (x, n (x), t (x)) $ as the matrix with first column $ x $, second column $ n (x) $ and third column $ t (x) $ – Danimenru May 15 '21 at 05:00
  • @VictorDanielMendozaRubio I have made an edit. – Arctic Char May 15 '21 at 11:08