Maps between Riemann surfaces are open and continuous I'm having some trouble with a couple of concepts in Riemman surfaces that I would really appreciate some help clarifying!
Firstly, is it true that a holomorphic map between two Riemann surfaces $f:R \to S$ is also continuous with respect to their topologies? I don't think this is as trivial as saying that holomorphic implies continuous because holomorphic is defined in terms of local coordinates. My attempt at a proof - but I'm really not sure this is correct:
Let $r \in R$ and say $s = f(r)$ then we have local coordinates $\phi: U_r \to \mathbb{C}$ and $\psi: U_s \to \mathbb{C}$ for some open sets around $r$ and $s$. Now from definition of holomorphic map between Riemann surfaces $g = \psi^{-1} f \phi$ is a holomorphic map $\mathbb{C} \to \mathbb{C}$ and as $\psi$, $\phi$ are invertible we have $f = \phi^{-1} g \psi$ is holomorphic and so $f$ is continuous?
Secondly I'm trying to prove the open mapping theorem. From the local form of a holomorphic map if we take an open set $U$ in $R$ then we can take a point $p \in U$ and looking locally we have $f(z) = z^n$, where $z = \phi (p)$ but is it obvious that $z^n$ is an open map? If so I'm not sure how to prove this?
 A: "are invertible we have $f=\phi ^{-1}g\psi$ is holomorphic and so f is continuous?" ---- Careful.  What $f$ actually denotes here is the restriction of $f$ to $U_r$, and so $f|_{U_r}$ is continuous, not $f$ itself.
But this fact allows you to then show that $f:R\rightarrow S$ is continuous.  For each $r\in R$, pick some $U_r$ as before.  Then, you've showed that $f|_{U_r}$ is continuous.  As $\left\{ U_r:r\in R\right\}$ covers $R$, it follows that $f$ is continuous.
Let $U$ be an open connected subset of $\mathbb{C}$ and define $f:U\rightarrow \mathbb{C}$ by $f(z):=z^n$ for $n\in \mathbb{Z}^+$.  We want to show that $f$ is open.  Let $V$ be any open subset of $U$.  You want to show that $f(V)$ is open, so cover $V$ by open balls $B_i:=\left\{ z\in V:|z-z_i|<r_i\right\}$.  As $f(V)=\bigcup _if(B_i)$, it suffices to show that each $f(B_i)$ is open.  However, by replacing $z^n$ with $(\tfrac{z-z_i}{r_i})^n$, it suffices to show that $f(D)$ is open where $D$ is the unit disk.  (Note that this requires an induction argument, because you need to use the fact that $z^k$ is open for $1\leq k\leq n-1$ in order for the open-ness of $z^n$ and $(\tfrac{z-z_i}{r_i})^n$ to be equivalent).  But
$$
f(D)=\left\{ z^n:|z|<1\right\} =D
$$
is open
A: It seems to me that the argument about continuity is correct, because continuity and holomorphicity are local properties. So, if $f|{U_r}$ is continuous for all $r$, then $f$ is itself continuous. (It is enough to check on an open cover)
The derivative of function $f(z)=z^n$ is non-zero at all $z\neq 0$. So for all neighbourhoods of $z\neq 0$ we can apply the inverse function theorem to say that $f$ is an open map.
So the only problematic point is $z=0$. To say that any open neighbourhood of $z=0$ maps to an open set, it is enough to say the same for open balls centered at $z=0$. 
Let $B=\{z\ |\ |z|<r\}$ for $r>0$. Then $f(B)=\{z\ |\ |z|<r^n\}$, which is open.
