Let $M(2, \Bbb C)$ be the set of all complex square matrices of degree 2, and define its subset X as follows. $X = \{ A ∈ M(2,\Bbb C)|A^2 = E, A≠ E, A ≠−E\}$, where $E$ is the second-order identity matrix. Gives $X$ the relative topology of $M(2, \Bbb C)$. Let $S^2 = \{(t, u) ∈ \Bbb R × \Bbb C | t ^2 + |u| ^2 = 1\}.$ For each $A ∈ X$,take $ α, β ∈ \Bbb C$ such that $A \left [ \begin{matrix} \alpha \\ \beta \\ \end{matrix} \right ] =\left [ \begin{matrix} \alpha \\ \beta \\ \end{matrix} \right ]$ and $|α|^2 + |β|^2 = 1$. Then set $f(A) = (|α| ^2 − |β|^ 2 , 2\bar\alpha\beta).$Prove that
(1) $f$ is a map from $X$ to $S^2$ that is independent of the choice of $\alpha$ and $\beta$.
(2) $f$ is a continuous map.
(3) For each point $p$ in $S^2$, the inverse image $f^{-1}(p)$ is homeomorphic to $R^2$.
I'm not sure the proof of $(1)$ I write is correct: $f(A)\in S^2$is clearly.The eigenvalue of $A$ could be $1$ or $-1$,since$ A≠ E, A ≠−E$,it must have eigenvalue both $1$ and $-1$,so if another $x=(\hat \alpha,\hat\beta)$satisfies$Ax=x$,then $x$ must be the multiplication of $(\alpha,\beta)$,i.e. $\hat \alpha=c\alpha,\hat\beta=c\beta$,the condiction$|α|^2 + |β|^2 = 1$ yields that $|c|^2=1$,then$|\hatα| ^2 − |\hatβ|^ 2=|c|^2 (|α| ^2 − |β|^ 2)=|α| ^2 − |β|^ 2 , 2\bar{\hat\alpha} \hat\beta=|c|^22\bar\alpha\beta=2\bar\alpha\beta.$ But for$(2)$and$(3)$ I have no idea so please help me,thank you!