Fundamental group of extended complex plane

Question

I am reading lecture notes of Riemann surfaces and encountering something I cannot understand.

As is shown in Example 4.5, the fundamental group of $\mathbb C^*$ is isomorphic to $\mathbb Z$. However, it implies that $\mathbb C^*$ is not simply connected, which contradicts

$$\pi_1(S^2)=\{0\}.$$

But I can't see where I go wrong. Any comments is appreciated:)

Answer 1

Your question was answered in the comments, but let me give an offcial answer to clear it from the "unanswered" queue.

The space $\mathbb C^*$ does not denote the Riemann sphere (which is homeomorphic to $S^2$), but the punctured complex plane $\mathbb C \setminus \{0\}$. This is homotopy equivalent to $S^1$, hence $\pi_1(\mathbb C^*) \cong \mathbb Z$.

Notation is explained in the linked paper:

Section Notation on page ii :

• We use $\mathbb C^* = \mathbb C \setminus \{0\}$

Page 1 :

• Example 1.3 (Riemann sphere). Let $\hat{\mathbb C} := \mathbb C ∪ \{∞\}$

But even without these explicit definitions it is clear from the context that $\mathbb C^*$ must denote the punctured complex plane. The author considers the universal covering $\exp : \mathbb C \to \mathbb C^*$. Coverings are surjective by definiton, and the image $\exp(\mathbb C)$ is precisely the punctured complex plane.