Cut and paste construction of Riemann surface of $z\mapsto \sqrt{z}$, the two edges of the cut I'm interested in trying to understand in detail the ''cut and paste'' construction of Riemann surfaces of simple analytical functions. I'm convinced that I understand the pasting part, but I'm not so clear on the cutting, in fact, it appears I'm confused about defining the ``edges'' we are going to paste along.
Let me be a bit more precise: suppose that we are cutting along the positive real axis $\mathbb{R}_+$ in the complex plane $\mathbb{C}$. Intuitively, we get two edges along the cut. My question is how to construct this new topological space in which we have the cut. In some sense one needs to complete $\mathbb{C}\setminus \mathbb{R}_+$ by adding two copies of $\mathbb{R}_+$ in such a way that one copy is the set of limit points of sequences approaching $\mathbb{R}_+$ with negative imaginary part, and the other the limit points of sequences approaching $\mathbb{R}_+$ with positive imaginary part.
 Maybe one way could be to take two copies of $\mathbb{C}$ and paste them together by identifying $\mathbb{C}\setminus \mathbb{R}_+$ in both copies? 
Edit: What I suggested here doesn't give the correct topology, because it doesn't distinguish between ''above'' and ''below'' the cut.
 A: Typically one would take two "copies" of $\Bbb C$ and consider a quotient space of their union. But another alternative is to actually do the cutting and pasting in a higher dimension. Let me start with a figure-8 in the plane. It's got a crossing, which I'll declare "bad" (imagine it's an auto-race-course: collisions would happen there!). But if we take one strand of the figure 8 and lift it up, and press down the other strand (along an axis perpendicular to the plane), we get an overpass/underpass situation where there are no intersections. I propose to do much the same thing for the Riemann surface you're trying to build. First, I'll regard $\Bbb C$ as the $xy$-plane of $xyst$-space. I'm going to describe an embedding of $K = \Bbb C - \Bbb R^+$ that's a little different from the standard one. I'll need the functions $r(x, y) = \sqrt{x^2 + y^2}$ and $\theta(x, y)$, which is basically "arctan", with the "cut" placed along the positive real axis, so that the codomain is $[0, 2\pi)$. When I use these, I'll just write "$r$" instead of $r(x, y)$, to make thing a bit less messy. 
Here's a first embedding of $K$:
$$
f_1 : K \to \Bbb R^4 : (x, y) \mapsto (x, y, \theta - \pi, 0).
$$
Here's a second embedding of $K$:
$$
f_1 : K \to \Bbb R^4 : (x, y) \mapsto (x, y, \pi - \theta, 0).
$$
If we let $A = f_1(K)$ and $B = f_2(K)$, then $A \cup B$ is almost the standard textbook picture of the Riemann surface of $z \mapsto \sqrt{z}$. (Yeah, there should be a $\sqrt{r}$ in there, but I'm trying to get the topology right, so I'm not doing every detail.)
To get the actual Riemann surface, I'm going to do two more things. First, the boundary of $f_1(K)$ lies entirely in the $y = 0$ plane, and consists of two half-lines joined at $(0,0,0,0)$; the same goes for $f_2(K)$. So if we take the union of the closures of $A$ and $B$, $\bar{A} \cup \bar{B}$, we get a surface without boundary. Unfortunately, it's also a surface with a self-intersection along the negative-$x$ axis (where the $s$ coordinate on each surface is zero). To fix this, we can define
$$
g_1(x, y) = f_1(x, y) + (0,0,0,r\sin^2 \theta) \\
g_2(x, y) = f_2(x, y) + (0,0,0,-r\sin^2 \theta) 
$$
The $\sin^2$ term acts like the "overpass/underpass" thing:
If $g_1(x,y) = g_2(x, y)$, then the $f$-values have to be equal, so we're on the negative-real axis (i.e., $x \le 0, y = 0$) on which $\theta(x, y) = \pi$, hence the $t$-coordinate for $g_1$ is positive, and the $t$-coordinate for $g_2$ is negative. The only tricky case is $x,y = 0$, where the $s$ and $t$ coordinates are the same, so the two points are actually identical. The set
$$
A' = g_1(K)\\
B' = g_2(K)
$$
share the origin. If we look at $\bar{A'} \cup \bar{B'}$, we actually get a surface that really is an embedding of the cut-and-paste model of your Riemann surface into $\Bbb R^4$. 
I don't know if this helps, but it was worth a shot. 
