Sufficient condition for an holomorphic map to be conformal Let $U,V\subseteq\Bbb C$ be open sets, let $f:U\to\Bbb C$ be holomorphic.
If we want to prove that $f$ is a conformal map $U\to V$, my teacher said that is enough to check that $f$ is locally conformal on $U$ (i.e. $f'$ never vanishes on $U$) and $f(\partial U)=\partial V$.
I think it works for simply connected open subsets, not for arbitrarily open subsets; to me this is a topological problem.
I'm quite disoriented: why is this true? I know that an holomorphic map is locally conformal iff is locally injective, but this does not imply global injectivity! How can be injectivity and surjectivity ensured by the condition on the boundary?
EDIT: by conformal, here I mean biholomorphic, i.e. holomorphic and bijective (the holomorphy of the the inverse follows from this).
 A: This does not hold in general. Consider $$f:\mathbb{D_*}\to \mathbb{D_*}$$$$f(z)=z^2$$ 
Clearly $f$ is locally conformal, and since $\partial \mathbb{D_*}=\mathbb{S^1}\cup\{0\}$, $f(\partial \mathbb{D_*})=\partial \mathbb{D_*}$. But $f$ is not globally conformal.
I assume you mean that $f$ extends continuously to $\partial U$, otherwise we can "patch" up the interior and the boundary in any which way. Although this isn't completely general, it does seem to be true for simply-connected domains $U$ and $V$ with locally connected boundaries. We appeal to an extension to the $\textbf{Caratheodory Extension Theorem}$, see here: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(conformal_mapping). 
We can replace $U$ and $V$ with $\mathbb{\overline D}$ by virtue of the theorem. Now any map $f$ continuous on $\mathbb{\overline D}$ and holomorphic on $\mathbb{D}$ satisfying $|f|=1$ on $|z|=1$ is a Blaschke product. A direct computation shows that a locally conformal Blaschke Product has only one factor, i.e, is a biholomorphism. This settles the question.
Yet another interesting case pops up. As @studiosus notes, the assumption that $f$ is proper gives a positive result. Indeed, any proper local biholomorphism is a covering action, and simple-connectivity of $V$ shows that $f$ has to be univalent, which is all that we need. 
