Question. Consider the function $y=\sqrt{x^4-x^2+1}$. Consider a sufficiently large circle $C$ of $0$ in $\mathbb{C}$, and let $x_0$ lies in this circle (suppose $x$ is not a branch point of this function). We choose a solution $y_0$ for $x_0$. How would one show that by analytically continueing $(x_0,y_0)$ along this circle, when one returns to $x_0$, the corresponding solution is still $y_0$, rather than $-y_0$?
Side note. In other words, consider the double covering $\varphi:\{(x,y): y^2=x^4-x^2+1\}\to \mathbb{C}$ by projecting to $x$. For sufficiently large $R$, $U_R:=\{z\in \mathbb{C}: |z|>R\}$, the above question is the same as asking that $\varphi^{-1}(U_R)$ is not path-connected.
Or equivalently, from $y^2=x^2-x^2+1$, we get a double covering map $X\to \mathbb{P}^1$ where $X$ is a compact Riemann surface. The above question is equivalently to asking to show that $\infty$ is unramified.
An attempt. I learned this from page 13, but I am not completely convinced by this informal approach: As $x$ goes around the circle and get back, its argument increases by $2\pi$, so the arguments of $x^4$ and $x^2$ increase by $8\pi$ and $4\pi$, respectively. It means the argument of $x^4-x^2$ increases by $8\pi-4\pi=4\pi$, meaning $x^4-x^2+1$ has argument increases by $4\pi$ (with respect to center $1$), so $y^2$ has argument increases by $4\pi$ around $1$ (I don't think this makes sense?), so $y$ has argument increases by $2\pi$. This means we cannot reach $-y$ by analytically continuing along this circle.
Any help would be much appreciated!
Here's a link, in case you don't know what it means for two paths to be homotopic:
https://en.wikipedia.org/wiki/Path_(topology)#:~:text=Homotopy%20of%20paths,-Main%20article%3A%20Homotopy&text=Paths%20and%20loops%20are%20central,while%20keeping%20its%20endpoints%20fixed.
Let $g(z)=z^4-z^2+1$.
The idea of the "informal approach" is that the path $g(Re^{2\pi it})$, where $t$ ranges from $0$ to $1$, is homotopic to the path $g(R)e^{8\pi it}$, that is to say the path that starts at $g(R)$ and goes round the circle clockwise four times at a fixed rate.
That will indeed allow you to construct a continuous square root of $g(z)$, at least for $z$ on the circle $C$.
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There is another, even simpler approach. For the sake of generality, I'll talk about finding the square root of $h(z)=p(z)$, where $p(z)$ is a polynomial and $\deg{p(z)}=2n$.
Think about $u(z)=z^{2n}h(1/z)$. For a small enough $r$, the square root of $u(z)$ is analytic on $|z|<r$. Call this square root $w(z)$.
Clearly $z^{-n}w(z)$ is a square root of $h(1/z)$ that is analytic on $0<|z|<r$.
So $z^{n}w(1/z)$ is a square root of $h(z)$ that is analytic on $|z|>1/r$.