# Branch points on Riemann surfaces

I have a question on the exact definition of branch points (on Riemann surfaces) I have the following definition::

Let $$f:X \rightarrow Y$$ be a holomorphic function between Riemann surfaces. $$x \in X$$ is a branch point, if the multiplicity, with which $$f$$ takes the value $$f(x)$$ in $$x$$ is $$>1$$. The mmultiplicity can be computed by expressing $$f$$ as a function between open susets of $$\mathbb{C}$$ by using charts.

In other words, branch points are the points where $$f$$ has no neighborhood on which it is injective.

For example, look at the function

$$f: \mathbb{P}^1 \rightarrow \mathbb{P}^1, z \mapsto z^2 + \dfrac{1}{z^2}$$

I computed the branch points to be $$1, 0, \infty$$ by constructing the function with the charts.

Now, I found this exercise on the internet, where they computed the points $$\pm 2$$ and $$\infty$$ as branch points.

Did I make a mistake or is there another definition for branch points?

Note that $$f(-z)=f(z), f(1/z)=f(z)$$ so certainly if $$z_0$$ is a branch point $$\pm 1/z_0, -z_0$$ are other branch points.
Also $$f(-z)=f(z)$$ implies that $$0$$ is a branch point, as around $$0$$ $$f(-a)=f(a)$$. Hence $$0, \infty$$ are branch points.
For seeking others, one can make stupid computations $$f'(z)=2z-{2\over z^3}$$, and $$f'(z)=0$$ iff $$z^4=1$$, ie $$z\in \{1,-1,i,-i\}$$. Conclusion 6 branch points$$\{0, \infty,1,-1,i,-i\}$$.
• No it is the same $f'(x_0) = 0$ iff $x_0$ is a branch point iff $f$ is not locally injective near $x_0$ ($f'(x_0)\not = 0$ implies that locally one can choose $f$ as a coordinate (implicit function theorem). – Thomas Feb 22 at 11:11
• oh no I didn't mean your definition, I get that, I meant the one I found where the branch points were $\pm 2$. – User1 Feb 22 at 11:28
• @User1 That is correct, they're probably using this definition. Then $0, \pm 1, \pm i, \infty$ are ramification points (each of index $2$), and the corresponding $f(z)$ gives three branch points $\pm 2, \infty$. – Maxim Mar 9 at 17:20