# Gluing together Riemann surfaces, don't see why $Z$ is union of two compact sets.

Consider Lemma 1.7 from page 60-61 of Miranda's Algebraic Curves and Riemann Surfaces. For a link to the book, see here.

Lemma 1.7. With the above construction, $$Z$$ is a compact surface of genus $$g$$. The meromorphic function $$x$$ on $$X$$ extends to a holomorphic map $$\pi: Z \to \mathbb{C}_\infty$$, which has degree $$2$$. The branch points of $$\pi$$ are the roots of $$h$$ (and the point $$\infty$$ if $$h$$ as odd degree).

Proof. One checks readily that $$Z$$ is Hausdorff, and hence is a Riemann surface. $$Z$$ is compact, since it is the union of two compact sets$$\{(x, y) \in X : \|x\| \le 1\} \text{ and }\{(z, w) \in Y : \|z\| \le 1\}.$$

To me, I don't see why $$Z$$ is the union of the above two compact sets. Can anyone clarify? Thanks.

The two given sets are indeed compact, since $h$ and $k$ are bounded in modulus on the indicated unit disks. Now the point is just that in the first coordinate, the gluing map is $z = \tfrac{1}{x}$, so the exterior of the first set (i.e. $\{ (x,y)\in \mathbb C^2 \, | \, ||x|| > 1 \}$ ) is mapped to $\{ (z,w) \in \mathbb C^2 \, | \, 0 < ||z|| < 1 \}$. Thus every point of $X$ with $||x|| > 1$ is mapped to a point of $Y$ with $||z|| < 1$, and these two compacta cover $Z$.