Confusion about Riemann surfaces I am studying Riemman surfaces of an algebraic functions and I have a slightly strange question (which has been tormenting me a lot). Im following the construction given in https://projecteuclid.org/download/pdf_1/euclid.tmna/1471875703:
I will call algebraic function a multivalueded function $f$ on complex plane such that exists $g_0, \ldots, g_{n}\in \mathbb C[z]$ such that equation
$$F(z,f(z)) :=  \sum_{i=0}^n g_i(z)f(z)^i = 0$$
has a solution $a\in \mathbb C$. By the implicit function theorem, it's easy to show that if $F$ has a $n$ different solution's $z_1,\ldots,z_n \in \mathbb C$ , $F(a,z_i) = 0$, then there is a disk $U_a$ in $a$ with $n$ functions $(f_{a,i})_{i\leqslant n}$ such that  $F(z,f_{a,i}(z)) = 0$ for all $z\in U_a$.
From above, there is a way to construct extensions of some branch $f_a$ of $f$ throught paths $\gamma$ (such that $\gamma(0)= a$). Monodromy theorem comes in play to show that these extensions are unique up to homotopic curves with the same endpoints.
Here troubles arises here: The author shouts that the union of all analytic elements $(f_a,U_a)$ forms the Riemann surface of $f$. Can i see that each ``sheet'' of the riemann surface corresponds to a copy of $\mathbb C-\{\text{singularities}\}$ where $f$ corresponds to some $f_a$? But am i running thought all $a\in \mathbb C$ such that $F(a,f(a))=0$ so... How that will help me to build the riemann surface of an algebraic equation? Sorry, many questions, but im quite crazy with that.
In all cases, thanks.
 A: First I suggest that you ignore the $f$ thing and instead consider the 2-variable polynomial
$$F(z,w) = \sum_{i=0}^n g_i(z) w^i
$$
The Riemann surface $S$ itself is then defined to be the solution set in $\mathbb C^2$ of the equation $F(z,w)=0$. One needs some assumptions on the polynomial $F$ including an irreducibility assumption so that $S$ is a connected surface (as opposed to something disconnected in which case the statement that you are asking about will be false).
The singular set is then a finite subset $\Sigma = \{(z_j,w_j)\}_1^J \subset S$, which projects to a finite subsets $A_z=\{z_j\}_1^J$ of the $z$-plane and $A_w=\{w_j\}_{1^J}$ of the $w$-plane, such that the projection maps $p_1(z,w)=z$ and $p_2(z,w)=w$ restrict to a pair of finite degree branched covering maps of $\Sigma$ over $\mathbb C$, where $A_z$ and $A_w$ are the respective branch sets, and so that the further restrictions $\pi_1 : S-\Sigma \to \mathbb C - A$ and $\pi_1 : S-\Sigma \to \mathbb C - B$ are actual finite degree covering maps. Also, the fact that $S$ is a connected surface implies that $S-\Sigma$ is also connected, in fact path connected.
So now as $x \in S-\Sigma$ varies, the pairs $(z,w)=(\pi_1(x),\pi_2(x))$ vary in exactly the way you describe. For example suppose you have $z_0 \in \mathbb C-A$ and two values $w_1,w_2 \in \mathbb C - B$ such that $F(z_0,w_1)=F(z_0,w_2)=0$. Using path connectivity of $S-\Sigma$, you choose a path $\gamma$ in $S-\Sigma$ from $(z_0,w_1)$ to $(z_0,w_2)$, and the projected path $\gamma_1 = p_1 \circ \gamma$ is a closed path based at $z_0$ which lifts to the chosen path $\gamma$ in $S-\Sigma$ which in turn projects to $\mathbb C-B$ to form a path from $w_1$ to $w_2$.
