If we have $y^2=(x-1)(x-2)\dots(x-2n)$, do the roots get exchanges if we run around the origin once? This is an example from the much-used Algebraic Geometry notes by Gathmann.

He says that if we consider the equation $y^2=(x-1)(x-2)\dots(x-2n)$, where $x,y\in\Bbb{C}$, then

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*The solution set is two complex planes with the points $\{1,2,3,\dots,2n\}$ identified.


*If we run around the origin once, the two solutions of $y$ get exchanged.


*

*I don't understand why this is the case. I can understand that for each value of $x$, there are two values of $y$. However, do the two complex planes denote the two values of $y$, or the values of $x$? Is it true that all that this diagram of two complex planes joined at the given points indicates is that the two values of $y$ co-incide at these points, and nothing else?


*Should the two values get exchanged even if $n$ is an even number? I mean if we run around the origin once, we're rotating each factor $(x-p)$ by an angle of $2\pi$. If $n$ is even, we're rotating the product of all factors by an angle of $4k\pi$. Shouldn't the values of $y$ remain the same then?
 A: *

*Your remark is correct, for all values of $x$ there are two values of $y$ that make $(x,y)$ a solution, except if $x=1, 2 \ldots$ or $2n$, in which case there is only $y=0$. Thus the complex curve of the solutions may be identified with 2 copies of the complex plane, which are parameterized by $x \in \mathbb C$. For a given $x$, the 2 corresponding points in each plane correspond to the 2 values of $y$ making $(x,y)$ a solution. Since there is only 1 solution in the case $x=1, 2, \ldots, 2n$, you need to glue those 2 planes together at those points.

*I suspect you or the author of the book meant "if you run around one of the $k$, $k=1, \ldots , 2n$" instead of "if you run around the origin". If you run around the origin in a loop contained in the unit disk so that it avoids the singularities located at $1, \ldots, 2n$, you don't switch the solutions. That's because you can define holomorphic branches $\pm \sqrt{ (x-1)\ldots (x-2n)}$ on the unit disk.
However, if you run around once along a small loop centered at one of the singularities, you will indeed switch the two solutions. Here's a way to see it elementarily on the loop $\gamma(t)=1+ 0.5 e^{it}$: you want to solve $y(t)^2 = (\gamma(t)-1) \ldots (\gamma(t)-2n)$, and let's say that you choose $\gamma(0)$ to be the one with positive imaginary part. Note that since $(x-2) \ldots (x-2n)$ doesn't vanish on the disk $\mathbb{D}(1,1)$, you can find a holomorphic function $\phi$ defined on $\mathbb{D}(1,1)$ such that $\phi(x)^2=(x-2)\ldots (x-2n)$ and $y(0)=\sqrt{ \frac{1}{2} } \phi(\gamma(0))$. So our equation boils down to $y(t)^2=(\gamma(t)-1) \phi(\gamma(t)) = 0.5 e^{it} \phi(\gamma(t)).$
So we find $y(t)=\sqrt{ \frac{1}{2} } e^{it/2} \phi( \gamma(t) )$, and as expected if we put in $t=2\pi$ we find $y(2\pi)=-\sqrt{ \frac{1}{2} } \phi(\gamma(0))=-y(0)$.
This phenomenon is known as monodromy. If you read about covers and fundamental groups, you will find a more general setting.
