# 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?

"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

• Thank you for this detailed answer. I have a small question. When showing $f$ is open, if we take an open set $V$ in $R$, and then I take a point $p \in V$ can we be sure that there is an open neighbourhood $U$ of $p$ that is contained in $V$, couldn't it be bigger than $V$? Once this is sorted, I think I understand the rest of the proof! – Wooster Nov 4 '14 at 20:28
• Of course. For example, $V$ itself is an open neighborhood of $p$ contained in $V$. Though I must admit I'm not exactly sure what in the proof made you ask this. – Jonathan Gleason Nov 4 '14 at 21:43
• Also, for what it's worth, I think I prefer poorna's argument that it we really need only consider balls centered at the origin. – Jonathan Gleason Nov 4 '14 at 21:44
• But don't we also need a homeomorphism $\phi: V \to \mathbb{C}$? And we are only given one on some $U_p$, not necessarily the same set as $V$? – Wooster Nov 4 '14 at 22:18
• It suffices to show that $f$ is open in a neighborhood of every point. Suppose we have shown this, that is, suppose we have shown that for every point $p\in R$, there is an open subset $V_p$ containing $p$ such that $f|_{V_p}$ is open. Now let $U$ be an arbitrary open set and define $U_p:=U\cap V_p$. Then, $f(U)=f(\bigcup _pU_p)=\bigcup _pf(U_p)$. We know that each $f(U_p)$ is open (because it is a subset of $V_p$ and $f$ restricted to $V_p$ is open), and hence $f(U)$ is. – Jonathan Gleason Nov 5 '14 at 22:09

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.