How to compute this Riemann surface? This question is related to other more general question that I asked Computing Riemann surfaces of a given algebraic function. By the way, I've found an approaching in Markushevich's book that satisfies in some way the question in the link. However I still not being able to solve some problems. 
For instance, let $$ z = \frac{1}{2}(w + \frac{1}{w}) $$ and $$z = \frac{1}{2}(w + \frac{1}{w^2}) $$
How do I compute the associated Riemann surfaces?
The general procedure is to find where the function $f(w)= z$ is injective around each branched point by picking some "triangles" with vertex in the respective branch point and, then, paste these "triangles" along the boundaries in the same way that the images are glued. 
For the first case, it seems that I need find a double cover since the first equation is non-injective for $w.w'=1 $, then $1$ and $-1$ would be the unique branch points, however I don't know how to find the "triangles" in the real axis (at first, I was thinking in picking the upper semi-plane and the lower one, however points in the real line when reflected stay in the real line). Now, for the second algebraic function, I have no idea about how to start.
Thanks in advance.
 A: It's not the topological approach you ask about, but one standard algebraic technique is to write your curve as the locus of a homogeneous polynomial. For your first curve, you have $2zw - w^2 - 1 = 0$; treating $z = Z/T$ and $w = W/T$ as ratios of homogeneous coordinates in $\mathbf{P}^2$, the preceding equation becomes $2ZW - W^2 - T^2 = 0$. This is a smooth conic (by the Implicit Function Theorem, for example), hence isomorphic to $\mathbf{P}^1$ by the degree-genus formula. To see what the curve looks like in the other two affine charts for $\mathbf{P}^2$, set $Z = 1$ (obtaining $2w - w^2 - t^2 = 0$ in affine coordinates $w = W/Z$ and $t = T/Z$) or $W = 1$ (obtaining $2z - 1 - t^2 = 0$).
By a similar procedure, your second curve has homogeneous equation $2ZW^2 - W^3 - T^3 = 0$. This is a singular cubic, with a node at $[T:Z:W] = [0:1:0]$. The affine representations in the other two charts are $2w^2 - w^3 - t^3 = 0$ (when $Z = 1$) and $2z - 1 - t^3 = 0$ (when $W = 1$).
Though they're not identical to your examples, the images here may be helpful in visualizing your surfaces.
Added in edit: An algebraic locus $F(z, w) = 0$ in $\mathbf{C}^2$ can be compactified in $\mathbf{P}^2$ by homogenization as described above, or in $\mathbf{P}^1 \times \mathbf{P}^1$ by introducing variables $u = 1/z$ and $v = 1/w$, considering the resulting four loci, and attaching them by identifying, e.g., $(z, w)$ and $(1/u, w)$.[*]
The "customary" topological construction of a Riemann surface in $\mathbf{P}^1 \times \mathbf{P}^1$ starts with a multi-valued function (a.k.a. branched covering), say $w$ expressed as a multi-valued function of $z$. One takes a "copy" of $\mathbf{C}$ for each sheet of the covering, makes appropriate cuts between branch points in each copy, then joins edges of slits consistently with analytic continuation.
In the two examples at hand, $z$ is already expressed as a meromorphic function of $w$, so the "slit sheets" construction isn't strictly necessary; each Riemann surface is biholomorphic to $\mathbf{P}^1$ via $w \mapsto (z, w)$. However, here's how the construction might look for the first, which is formally simpler to handle:
The locus $z = \frac{1}{2}(w + \frac{1}{w}) = f(w)$ may be written $w^2 - 2zw + 1 = 0$. Using the quadratic formula to solve for $w$, we have $w = -z \pm \sqrt{z^2 - 1}$, a $2$-valued function with branch points $z = \pm 1$ (the critical values of $f$). To construct the Riemann surface, fix a branch of the square root, say, the branch defined off the non-positive real axis that is positive on the positive real axis. Take two copies of $\mathbf{C}$, slit each along the segment $[-1, 1]$ where $\sqrt{z^2 - 1}$ is undefined, and "cross-join" the edges of the slits, so that (e.g.) the positive imaginary axis in one copy "continues" to the negative imaginary axis in the other copy. There is no branching at $\infty$, so the resulting surface may be visualized as a pair of slit ping-pong balls with the slits cross-glued; this surface is topologically a sphere (a copy of $\mathbf{P}^1$ with coordinate $w$), and the two-sheeted covering is effected by the mapping $f$, whose image is a copy of $\mathbf{P}^1$ with coordinate $z$.
One "classical" construction of a torus begins with an equation of the form $z^2 = w(w - 1)(w - w_0)$, with $w_0$ a complex number different from $0$ and $1$. Without going into great detail, one takes two copies of $\mathbf{C}$ and makes the same two slits in each so that each branch point $0$, $1$, $w_0$, and $\infty$ is an end of a slit. Again, cross-glue the edges of the slits. Topologically, two spheres have been joined by removing two "corresponding" disks from each and attaching the corresponding boundary circles. By taking a polynomial of degree $2g+1$ or $2g+2$ on the right, one similarly obtains a Riemann surface of arbitrary genus $g \geq 0$. (These surfaces are "hyperelliptic", i.e., they admit a two-sheeted holomorphic map to $\mathbf{P}^1$, but at least it's possible to see how more complicated topological types arise.)
[*] To pass geometrically from one compactification to the other, let $[Z:W:T]$ denote homogeneous coordinates in $\mathbf{P}^2$; blow up the points $[1:0:0]$ and $[0:1:0]$, then blow down the proper transform of the line $T = 0$ "at infinity" in the affine chart with coordinates $(z, w) = (Z/T, W/T)$. To see this has the asserted effect, recall that a smooth quadric surface $Q$ in $\mathbf{P}^3$ is biholomorphic to $\mathbf{P}^1 \times \mathbf{P}^1$. Fix a point $p$ on the quadric and a hyperplane $P$ not passing through $p$. Projection away from $p$ maps $Q\setminus\{p\}$ to $P\setminus\{p\}$; this map collapses the two rulings of $Q$ passing through $p$, but is otherwise a biholomorphism, since a generic line in $\mathbf{P}^3$ hits the quadric $Q$ exactly twice. This projection does not extend holomorphically to $p$, but does extend to the blow-up of $Q$ at $p$. Consequently, the blow-up of $Q$ at $p$ is biholomorphic to the blow-up of $P$ at two points.
Second edit: Technically, the "cut-and-cross-glue" construction with two sheets works like this: Let $\mathbf{C}_1^+$ and $\mathbf{C}_1^-$ denote the closed upper and lower half-planes in the "first copy", and similarly for $\mathbf{C}_2^\pm$.
If $z_{0}$ is a point of $(-1,1)$ in $\mathbf{C}_1^+$, a disk neighborhood of $z_0$ is obtained from open half-disks of equal radius in $\mathbf{C}_1^+$ and $\mathbf{C}_2^-$ by identifying their boundary points. Similarly, if $z_{0}$ is a point of $(-1,1)$ in $\mathbf{C}_2^+$, a disk neighborhood of $z_0$ is obtained from open half-disks of equal radius in $\mathbf{C}_2^+$ and $\mathbf{C}_1^-$ by identifying their boundary points.
If $z_{0}$ is a point of $(-\infty,-1)$ or $(1, \infty)$ in $\mathbf{C}_1^+$, a disk neighborhood of $z_0$ is obtained from open half-disks of equal radius in $\mathbf{C}_1^+$ and $\mathbf{C}_1^-$ by identifying their boundary points, i.e., by rejoining what has been split asunder.
Finally, a disk neighborhood of $1$ (or of $-1$) is obtained from four half-disks of equal radius (smaller than $2$), one in each closed half-plane. The four centers are all identified. Each of the two boundary radii in each half-disk is identified with a boundary radius in some other half-disk, as shown:

Each real point $z_{0} \neq \pm 1$ corresponds to two points of the Riemann surface; the branch points $\pm 1$ correspond to one geometric point, but counted twice algebraically.
