$C$ denotes complex plane with 1 point compactification. It is clear that $\sqrt{(z-a)(z-b)}$ has a branch cut $[a,b]$ for $a\neq b$ by considering $z_1^2=(z_2-a)(z_2-b)$. I could imagine for $a\neq b$ that I get a torus.
Suppose now $a=b=0$. I want to study the topology of $(C\cup\{\infty\})^2$. $\sqrt{z^2}$ does not have branch point at $z=0$ as if I go around $z=0$ once, I do not see an extra minus sign picked up. According to gathmann's notes, "we can view this as a surface with one hangle less with two points identified."
According to my previous statement for $a=b=0$, I expect to get complex sphere instead. What have I done wrong here?
Clarification of second paragraph. The following is the original statement in Gathmann Algebraic Geometry Notes 2014 version Example 0.2 in Introduction chapter. (http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2014/chapter-0.pdf)
"Consider $\{(x_1,x_2)\in C^2:x_2^2=f(x_1)\}\subset C^2$ where $f$ is a degree $2n$ polynomial. If two roots of $f$ approaching to each other and finally coincide, this has the effect of shrinking one of the tubes connecting two planes until it reduces to a "singular point"(also called a node)...
Obviously, we can view this as a surface with one handle less, where in addition we identify two of points. Note that we can still see the handles when we draw the surface like this, just like one of the handles results from gluing two points."
Your second paragraph is quite unclear, so I will only say that you are correct, the Riemann surface of the equation $z_1^2=z_2^2$ is indeed a sphere. You can see this by considering the map $f(z)=z^2$ on the one point compactification $\mathbb{C} \cup \infty$, which is two-to-one except over $f(z)=0$ and over $f(z)=\infty$ where it is one-to-one.