Regarding how to prove a map is a quotient map [closed]

consider such map $$P:C^2\setminus{(0,0)} \to U =\{{B \in M_2(C)|B = B^*, B^2 = B, tr(B)=1 \}}$$ by $$P(z_0,z_1):= \frac{1}{\|z_0\|^2+\|z_1\|^2} \begin{pmatrix} \|z_0\|^2 \ \bar{z_0}z_1 \\ \bar{z_1}z_0\ \|z_1\|^2\end{pmatrix}$$

We wish to investigate whether this map is a quotient map. Now, it is possible to give a proof in the following manner:

We descend $$P:C^2\setminus{(0,0)}$$ to $$CP^1$$ and applied the result $$CP^1 \cong M \cong S^2$$, which will force $$P$$ be a quotient map.

However it is possible to directly prove it? (for example, by proving the map is an open map)

I suspect that $$U$$ belong to some special group and we can exploit such property, but I don't really know how to proceed.

*any comment/idea will be welcome, I lack basic knowledge in Lie Algebra and Algebraic Geometry, and I only require some inspiration to work out the full problem.

• Your first method seems to be using the false premisse that if two spaces are homeomorphic then any map between them is a homeomorphism.
– Ruy
May 14 at 3:55
• Do I understand correctly that you have a proof, but want to have a more direct proof? May 14 at 23:49
• Have you proved that $P$ is a surjection? May 15 at 16:27

Consider the equivalence relation on $$\mathbb C^2\setminus \{0\}$$ given by $$x\sim y$$ iff $$P(x)=P(y)$$, so you have a continuous, bijective map $$\pi:\frac{\mathbb C^2\setminus \{0\}}\sim\to U.$$
Next you should prove that the LHS is compact, and that $$U$$ is Hausdorff, which would imply that $$\pi$$ is a homeomorphism, as desired.