Can a plane with n punctures be considered a riemann surface?

Question

I have read a result saying that the fundamental group for a Riemann surface of genus n is a set of 2n generators $a_i, b_i$ such that $a_1 b_1 a_2 b_2... a_n b_n a_1 ^{-1}b_1 ^{-1}...a_n^{-1} b_n^{-1}=1$. However, this is not the case for the fundamental group of an n-punctured plane, whose fundamental group is just the free group. But it seems to me that the punctured plane is a Riemann surface, since I can construct the mapping $\Pi_i 1/(z-z_i)$, $z_i$ being the holes, from it to the complex numbers. What am I missing? Does this have to do with completing the plane with a point at infinity?

Thank you!

Answer 1

The standard definition of a Riemann surface is a connected second-countable Hausdorff space with a complex atlas. The $n$-punctured plane qualifies. To quote Farkas & Kra, Riemann Surfaces:

We begin with a formal definition of a Riemann surface and give the simplest examples: the complex plane $\mathbb{C}$, the extended complex plane or Riemann sphere $\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$, and finally any open connected subset of a Riemann surface [my emphasis].

Or Miranda, Algebraic Curves and Riemann Surfaces:

Example 1.21. Any connected open subset of a Riemann surface is a Riemann surface; use the atlas on the subset as described in Example 1.15.

The issue in the rest of your question is that the phrase "a Riemann surface of genus $n$" usually implicitly assumes that it's a compact Riemann surface. Some authors define genus only for the compact case. For example, a section heading in Miranda: "The Genus of a Compact Riemann Surface".