Holomorphic function is local homeomorphism when its derivative is never zero Let $f:U\to\mathbb{C}$ be a non-constant holomorphic function, with $U\subset\mathbb{C}$ an open set. I want to prove that if $f$ verifies $f'(z)\neq 0 \ \forall z\in U$, then $f$ is a local homeomorphism. I would also like to know if the reciprocal statement is true, as  well as if there is an analogue statement when $f$ is an holomorphic function between Riemann surfaces.
I am assuming this claim is true, as I have seen it in several sites (one of them being the Wikipedia page for Local Homeomorphisms) and cited in some other proofs. Since an holomorphic function can be written as $$f(z) = \sum_{k=0}^{\infty}a_k(z-z_0)^k$$ in a certain open neighbourhood of $z_0$, the hypothesis tells us that $a_1\neq 0$, but I don't know where to go with this fact.
 A: It is true that a holomorphic function $f$ is locally injective (and locally biholomorphic) iff $f'(z)\neq 0$ for all $z\in U$.
For the $\Rightarrow)$ part, if we suppose $f'(z_0)=0$ for some $z_0\in U$ then locally, near $z_0$, we can write $f(z)-f(z_0)=(z-z_0)^k g(z)$ where $k\geq 2$ and $g$ is a holomorphic function such that $g(z_0)\neq 0$ (for this we only factorize $(z-z_0)^k$ in the Taylor series of $f(z)-f(z_0)$, note that this function is also injective near $z_0$). In particular, since $g$ is nonvanishing near $z_0$, there exists a function $h$ such that $h^k=g$, then $f(z)-f(z_0)=(zh(z))^k$ near $z_0$. Now by the open mapping theorem $zh(z)$ maps a neighborhood of $z_0$ onto an open neighborhood of $w=0$ in the range of $f-f(z_0)$, then $(zh(z))^k$ can not be injective since $k\geq 2$ and hence there exist at least two values $u,v$ being mapped into the same point (think about $z^2$ when is applied to the unit ball).
The $\Leftarrow)$ part also can be proved by means of the argument principle. Given $z_0\in U$ there exists an open ball $B(z_0,r)$ contained in $U$ and such that $f(z)-f(z_0)\neq 0$ for all $z\neq z_0$ in this ball, then
$$\frac{1}{2\pi i} \int_{\partial B(z_0,r)} \frac{f'(z)}{f(z)-w} \,dz,$$
in well defined for $w\in B(f(z_0),r_2)$ where $r_2$ is sufficiently small such that $w\not \in \partial f(B(z_0,r))$, and the integral varies continuously in $w$ giving always the number of zeros of $f(z)-w$ in $B(z_0,r)$, since this number is an integer for all $w$, and for $w=z_0$ is equal to $1$, it follows that $f$ is injective restricted to $f^{-1}(B(f(z_0),r_2))$.
To do the last part by using the inverse function theorem, identify $\mathbb{C}$ with $\mathbb{R}^2$ and think of $f$ as being defined in the euclidean plane. We know that $f$ is differentiable in the real sense, and if it's derivative is non zero then it's Jacobian matrix is an element of $SO(2)$ multiplied by a constant factor (remember the Cauchy-Riemann equations), taking a adequate neighborhood we can apply the inverse function theorem of calculus and get that $f^{-1}$ is also differentiable in the real sense, but it's Jacobian matrix must be also an element of $SO(2)$ multiplied by a constant, since $SO(2)$ is a group. Hence $f^{-1}$ also satisfies the Cauchy-Riemann equations.
I know that you asked for $f$ only to be a homeomorphism, but as we have seen a holomorphic function which is locally injective is locally biholomorphic and hence also a local homeomorphism.
This result can also be generalized to Riemann surfaces in the sense that always exist charts $(\phi,U)$ and $(\psi,V)$ with $a\in U$ and $f(a)\in V$, such that $\psi \circ f\circ \phi^{-1}(z) = z^k$ for some $k\geq 1$, when $f$ is a non constant function between given Riemann surfaces. The proof is somewhat analogous.
