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I am learning to understand the behavior of holomorphic mappings from open unit disk to itself. And I found a related question said that a holomorphic function is always conformal, while another related question said that conformal mapping from unit disk to itself is always bijective.

Now after the reading I'm a bit confused about the relationship among holomorphic mappings, conformal mappings and bijective holomorphic mappings from open unit disk to itself... Are they all equivalent, or does there exist a holomorphic mapping from open unit disk to itself that is not bijective?

J.L.
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If $f(z)=\frac{z^2}2$, then $f$ is neither an injective nor a surjective map from the open unit disk into itself.

José Carlos Santos
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  • Thanks for the example. In that case can I conclude that not all holomorphic mappings from open unit disk to itself is conformal? – J.L. Jun 07 '21 at 18:25
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    No. You seem to think that conformal maps are always bijective. That's not true. The map $z\mapsto\frac z2$ (from the open unit disk into itself) is conformal, but it's not bijective. – José Carlos Santos Jun 07 '21 at 18:32
  • Sorry if I misunderstood... Does the 'no' mean that all holomorphic mappings from open unit disk to itself are conformal? – J.L. Jun 07 '21 at 19:03
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    A conformal map is a holomorphic map $f$ such that $f'$ is never $0$. So, for instance, the map$$\begin{array}{ccc}\Bbb D&\longrightarrow&\Bbb D\z&\mapsto&z^2\end{array}$$is not conformal. – José Carlos Santos Jun 07 '21 at 19:08
  • Got it, thanks. – J.L. Jun 07 '21 at 19:09