# A simple example from Gathmann's Algebraic Geometry

I am studying Gathmann's algebraic geometry, and the very first example has left me absolutely baffled. In Example 0.1.1, he considers the equation $$C_n = \big\{(x,y) \in \mathbb{C}^2 : y^2 = (x-1)(x-2) \cdots(x - 2n)\big\} \subseteq \mathbb{C}^2$$ He explains that this is not given by taking two copies of $$\mathbb{C}$$ and pinching the points corresponding to the value $$1,\ldots,2n$$. He argues this by considering a path $$x = re^{i \phi}$$ and let $$\phi$$ run from $$0$$ to $$2\pi$$.

Fair enough, but then out of nowhere comes the conclusion: "The way to draw this topologically is to cut the two planes along the lines $$[1,2],\ldots,[2n-1,2n]$$, and to glue the two planes along these lines as in this picture. I absolutely fail to see how this image connects up to the equation. If I were to take the loop with $$r = 2.5$$ on the upper plane of the left picture, I'd never switch between the two planes!

What am I misunderstanding?

N.B. I am aware of this question, which corrects some misleading points about Gathmann's example. But it is still not at all clear to me why the picture is the way it is.

• What he should be saying, I think, is that the root changes if your loop passes through one of the branch cuts (deleted intervals). So $r=1.5$ works, but not $r=2.5$. I also don't think the linked question is entirely correct (specifically regarding the identifications). You do indeed glue A to A, B to B, and so on. If we start on the lower sheet and pass through $[1,2]$ from below, we jump up to the upper sheet. Continuing to wind around $1$, we pass through the interval again which causes us to jump back to the lower sheet; so after a revolution of $4\pi$ we get back to where we started. Commented Nov 3, 2020 at 1:31
• This does actually result in the same topological picture as the right hand side of the graphic, it's just trickier to visualize (you may need to do some additional mental cut-and-past). Commented Nov 3, 2020 at 1:32