# Confusion about definition of ramification points in Forster

In the book "Riemann surfaces" by Forster a ramification point of $f:Y\to X$ is defined as a point $y\in Y$ such that there is no neighborhood $V$ of $y$ such that $f|_V$ is injective. On the other hand given that $X$ and $Y$ are compact Riemann surfaces and $f$ holomorphic, if for an open set $V\subset Y$ the map $f:V\to f(V)$ is injective (hence bijective), then the degree of $f$ being equal to one on an open set, would force $degf=1$ and hence $f$ is an isomorphism. So by definition of ramification point in Forster, one could imply that any holomorphic map between compact Riemann surfaces is isomorphism (take a point $y$ which is not a ramification point then from the definition it follows that there exists a neighborhood $V\subset Y$ of $y$ such taht $f|_V:V\to V$ is bijective and then apply the above degree argument) which is of course non-sense. I don't know where exactly my mistake is but I believe that I am making a mistake somewhere and don not see where.

• The degree of a map is something global. Take $f\colon \widehat{\mathbb{C}} \to \widehat{\mathbb{C}}; \; z \mapsto z^n$ for $\lvert n\rvert > 1$ to see that there are non-constant holomorphic maps between compact Riemann surfaces that are not isomorphisms. Yet, except at $0$ and $\infty$, these maps are of course locally injective. The degree is in some sense "local on the range"; if $\operatorname{card} f^{-1}(x)$ is equal to $n$ on some (punctured) neighbourhood of $x\in X$, then the map has degree $n$, but you are looking at a neighbourhood of an $y\in Y$ in the domain. – Daniel Fischer Jun 21 '15 at 9:36

...then the degree of $f$ being equal to one on an open set, would force $\deg f = 1$ and hence f is an isomorphism.
As you wrote you have a neighborhood $V$ in which $f: V \to f(V)$ is injective and somehow you wrongly concluded that $f : Y \to X$ must be injective. Have a look here to get more insight: https://en.wikipedia.org/wiki/Covering_space