Finding the Riemann surface of $w = z^{1/2}$

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I'm trying to find the Euler characteristic of $R = \{(z,w) : f(w,z) = w^2 - z = 0\}$.

To do this I'm using the Riemann-Hurwitz theorem with the projection $\Pi: R \to \mathbb{C}P^1$. Now in local coordinates we can write this map as $\pi (w) = g(w)$ for some holomorphic $g$ with $f(g(w),w) = 0$ (This is just the implicit function theorem).

We can then see there is a ramification point at $(z,w) = (0,0)$ and the degree of the map is 2 and has branching index 1.

Now if I try and use the Riemman-Hurwitz formula I get $\chi(R) = 2\deg(f) - b(f)$ - but we need to compactify R first.

My question is: I know that $\chi(R)$ should be even and as the degree of $f$ is 2 I must have to only add one point at $\infty$ to $R$ so the branching index becomes 2. However is there an intuitive way to see why I should add only one point to $R$ to compactify it - perhaps by looking at solutions to $f(w,z) = 0$ for large $w$ and $z$ and secondly how do we decide what local coordinate to use near $\infty$.

Thanks

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In the local coordinates $z' = 1/z$ and $w' = 1/w$ near $\infty$, i.e., thinking of your Riemann surface as living in $\mathbf{P}^{1} \times \mathbf{P}^{1}$, the squaring map is $w' \mapsto (w')^{2} = 1/w^{2} = 1/z = z'$; that is, the branching behavior at $\infty$ is the same as the branching behavior at $0$.