Studying the neighborhoods of a point Problem:
Let be $X=\{(x,y,z) \in \mathbb{C}^3: z^2=xy \}$.
Prove that cannot exists a neighborhood of $(0,0,0)$ in $X$ which is homeomorphic to $D \times D$ where $D$ denote the unitary open disc of $\mathbb{C}$.
My attempt:
The suggestion of the exercise is to first prove that the map
$$p : \mathbb{C}^2 \to X$$
such that $p(s,t)=(s^2,t^2,ts)$ induces a covering $p: \mathbb{C}^2-\{0\} \to X-\{0\}$.
I prove it but I cannot manage to use it to conclude.
 A: By contradiction, suppose there exists $U$ neighborhood of $(0,0,0)\in X$ homeomorphic to $D\times D \cong \mathbb C^2$ by the map $\varphi:U\rightarrow \mathbb C^2$ and $\varphi(0,0,0)=(0,0)$.
We have that $U\setminus \{(0,0,0)\}$ is connected and locally path connected because it is homeomorphic to $\mathbb C^2\setminus \{(0,0,0)\}$. Hence the restriction:
$$
p:p^{-1}(U \setminus \{(0,0,0)\}) \rightarrow U\setminus \{(0,0,0)\}
$$
is a covering
Consider now the composition:
$$
p^{-1}(U \setminus \{(0,0,0)\}) \xrightarrow{p} U\setminus \{(0,0,0)\} \xrightarrow{\psi \ = \ \varphi_{|_{U\setminus \{(0,0,0)\} }}} \mathbb C^2\setminus \{(0,0)\}
$$
Since $p$ is a covering and $\psi$ is an homeomorphism, then $\psi\circ p$ is a covering of $\mathbb C^2\setminus \{(0,0)\}$.
But $\mathbb C^2\setminus \{(0,0)\}$ is simply connected and $p^{-1}(U \setminus \{(0,0,0)\})$ is path connected, then $\psi\circ p$ is an homeomorphism, and this is a contraddiction because $p$ is not injective in an neighborhood of $(0,0)$.
