How can I verify the image and and the fibers of this map? Let $B = \lbrace (V,W) \in \Bbb C^2 : \vert V \vert ^2 + \vert W \vert ^2 = 1 \rbrace.$
Let $h_2(\Bbb C)$ denote the set of $2 \times 2$ Hermitian matrices with entries in $\Bbb C$.
Let $q \colon B \rightarrow h_2(\Bbb C)$ send each $(V,W) \in B$ to the matrix
$$\left[ \begin{matrix} \vert V \vert^2 & V\overline{W} \\ W \overline{V} & \vert W \vert^2 \end{matrix} \right].$$
I want to show that the image of $q$ is topologically $S^2$ and that $q$ is a fibration with fibers $S^1$.
What I think will help so far:
If $\vert V \vert = 0$ or $\vert V \vert =1$, the image of any $(V,W) \in B$ will be 
$$\left[ \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right] \text{ or } \left[ \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right], \text{ respectively}.$$
If we fix $V \in \Bbb C$ with $0 < \vert V \vert < 1$, we know that any point in $B$ with first coordinate $V$ will have a second coordinate whose squared norm is $1-\vert V \vert^2$.  Specifying the argument of the second coordinate fully determines the point's image by $q$.
Suppose I have a matrix $M$ in the image of $q$, say $q(V,W) = M$.  Then $\vert V \vert^2 = M_{11}$ and $\vert W \vert^2 = M_{22}$.  
We know the value for $V\overline{W}$, namely $M_{12}$.  
If we choose a value for the argument of one of $V,W$, the argument of its counterpart is forced to satisfy Arg$(V)$ -  Arg$(W)$ = Arg$(M_{12})$.
What I want to eventually do is generalize this map so that it acts on the quaternions and the octonions.  In these cases, $q$ will be a fibration with image $S^4$, $S^8$, and fibers $S^3$, $S^7$, respectively.  These maps are Hopf maps.
Could someone help me with a proof that might allow this generalization?
 A: Regarding the fibers: Well, it seems relevant to check that your map doesn't have any critical points. I mean, sounds like something you probably have already done, and if not, I think you should. Assuming that there are no critical points i.e. the map is a submersion, all fibers are diffeomorphic to one another, so you only need to verify that one of the fibers is a circle. That's not too hard.
Regarding the image: Take the cylinder $C=[0,1]\times S^1$, and define a map $\varphi:C\to h_2(\mathbb{C})$ by
$$(u,\theta)\mapsto\left(\begin{array}{cc}u^2&u\sqrt{1-u^2}e^{i\theta}\\u\sqrt{1-u^2}e^{-i\theta}&1-u^2\end{array}\right).$$ It is easy to verify that the image of the above $q$ is equal to that of $\varphi$, so we want to show that $\varphi$ maps $C$ onto a 2-sphere. Note that $\varphi$ maps the whole left circle $\{0\}\times S^1$ to a single point, and the right circle $\{1\}\times S^1$ to another point, thus it induces a map from the quotient $C/\tilde{}$ to $h_2(\mathbb{C})$, where $\tilde{}$ is the equivalence relation that glues each of the mentioned circles to a point. Note that $C/\tilde{}$ is topologically $S^2$. Furthermore, $\varphi$ is injective everywhere else, and obviously continuous, so the image is indeed homeomorphic to $S^2$.
A: Let $\Bbb D$ be a real normed division algebra of dimension $n \in \lbrace 1, 2, 4, 8 \rbrace$.
Let $k \in [0, 1]$, and let $E_k = \lbrace V \overline{W} \in \Bbb D: \Vert V \Vert^2 = k \wedge \Vert W \Vert^2 = 1 - k \rbrace$.
If $k=0$ or $k=1$, $E_k = \lbrace 0 \rbrace.$
If $k \in (0,1)$, $E_k$ is homeomorphic to $\big\lbrace U \in \Bbb D : \Vert U \Vert = \sqrt{k-k^2} > 0 \big\rbrace \approx S^{n-1}$.
\begin{align*} \text{We have Im}(q) &= \lbrace q(V,W) : (V,W) \in B \rbrace \\&= \Bigg\lbrace \left[ \begin{matrix} \Vert V \Vert ^2 & V \overline{W} \\ W \overline{V} & \Vert W \Vert ^2 \end{matrix} \right] \in h_2(\Bbb D): \Vert V \Vert^2 + \Vert W \Vert^2 = 1 \Bigg\rbrace \\& \approx \big\lbrace (\Vert V \Vert, V\overline{W}) \in [0,1] \times \Bbb D: \Vert V \Vert^2 + \Vert W \Vert^2 = 1 \big\rbrace \\ &\approx \lbrace E_k : k \in [0,1] \rbrace \approx S^n.
\end{align*}
Suppose we have a matrix $M := \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \in$ Im$(q)$.
Suppose $a = 0$.  We know that $q(V,W) = M$ for some $(V,W) \in \partial P'$.
We must have $V = 0$, $b = V\overline{W} = 0$, $c = \overline{W}V = 0$, and $d=\Vert W \Vert^2 = 1 - \Vert V \Vert^2 = 1$.
Thus $q^{-1}(M) \approx \lbrace W \in \Bbb D : \Vert W \Vert =1 \rbrace \approx S^{n-1}$.  Now suppose $a \neq 0$.
The map $\pi\colon q^{-1}(M) \rightarrow \lbrace V \in \Bbb D : \Vert V \Vert = \sqrt{a}\rbrace$ which sends $(V,W) \mapsto V$ is a homeomorphism; I will justify its injectivity.
Suppose $V \in \text{ Im}(\pi),$ i.e. $\exists$ $W \in \Bbb D$ such that $(V,W) \in q^{-1}(M)$.
Then $W\overline{V} = c$, i.e. $W = c{\overline{V}}^{-1},$ and the preimage of $V$ is unique.
Since $\lbrace V \in \Bbb D : \Vert V \Vert = \sqrt{a}\rbrace \approx S^{n-1}$, we have $q^{-1}(M) \approx S^{n-1}$.
