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I have just came upon the following example. Consider $\pi : M \rightarrow \mathbb C$ the projection on the first coordinate, where

$$M = \mathbb{C}^{2} \setminus (\{(z,w) \in \mathbb{C}^{2} : w=0\} \cup \{(z,w) \in \mathbb{C}^{2} : z=w\} \cup \{(0,1)\}).$$

The example is taken from a question discussed here When is a holomorphic submersion with isomorphic fibers locally trivial?

It is clear that this is a submersion and that the fibers are biholomorphic to $\mathbb{C} \setminus \{0,1\}$. But is this locally trivial? The answer suggested there is a little bit confusing to me: suppose $\pi$ is locally trivial, then extend $\pi$ into the single point $(0,1)$, and then restrict to the fiber. So $\mathbb{C} \setminus \{0\}$ should be mapped onto $\mathbb{C} \setminus \{0,1\}$, contradiction as the latter is hyperbolic.

Can anyone explain a little bit?

Since I was not happy with the previos proof, I have tried my own. Suppose $\pi$ is locally trivial, so $\forall z \in \mathbb{C}$, there exist $U \ni z$ and $\varphi : \pi^{-1}(U) \rightarrow U \times (\mathbb{C} \setminus \{0,1\})$ that is a biholomorphic map.

If $z \neq 0$, then we restrict to the fiber above $z$ and we get that $\varphi_{z} : \mathbb{C} \setminus \{0,z\} \rightarrow \mathbb{C} \setminus \{0,1\}$ is also a biholomorphic map. Now look at $\varphi_{z} \circ F_{z} : \mathbb{C} \setminus \{0,1\} \rightarrow \mathbb{C} \setminus \{0,1\}$, where $F_{z}$ is the mutiplication by $z$. But the number of such maps is finite. Using the fact that $\varphi$ is continuous and that $U \setminus \{0\}$ is connected, we can assume that $\varphi_{z} \circ F_{z}$ is exactly the identity map.

Take a sequence $a_{n}=(\frac{1}{n},2)$ which converges to $(0,2)$. But $\varphi(\frac{1}{n},2)=(\frac{1}{n},2n)$ does not converge to $\varphi(0,2)$ which is $(0,\beta)$, for some $\beta \in \mathbb{C} \setminus \{0,1\}$; contradiction to $\varphi$ being continuous.

Is my proof correct?

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