# Is a smooth Riemann surface??

Let be $M$,$N$ and $X$ compact riemann surfaces, $f:M\to N$ an holomorphic map with ramifications points $\{p_1,\ldots,p_r\}$ with multiplicities $m_1,\ldots,m_r$ and, $g:X\to N$ holomorphic maps $\{q_1,\ldots,q_s\}$ with multiplicities $n_1,\ldots,n_s$. Now, let be $$Y\subset M\times X,~~~Y=\{(z,x)\in M\times X |~~f(z)=g(x) \}$$

My question: When $Y$ is a smooth riemann surface?

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If $(x, y) \in M\times X$, then $y \in X$ so what do you mean by $f(y)$ given that $f : M \to X$? – Michael Albanese Dec 17 '12 at 23:23
In fact $f(y)=g(x)$ – user52188 Dec 17 '12 at 23:32
Now you have $f(y)$ with $y \in X$ but $f : M \to X$ and $g(x)$ with $x \in M$ but $g : N \to X$. Do you mean $Y = \{(x, y) \in M\times N\ |\ f(x) = g(y)\}$? – Michael Albanese Dec 17 '12 at 23:34
my apologies for the mess – user52188 Dec 17 '12 at 23:50

Let $(z_0,x_0) \in Y$ be a point, and suppose that the map $f$ has $m$-fold ramification at $z_0$, and that $g$ has $n$-fold ramification at $x_0$. Then we may choose a local coordinates $t$ at $z_0$ and $u$ at $x_0$, and a local coordinate $v$ at the point $f(z_0) = g(x_0)$, so that, in terms of the coordinate $v$, the function $f$ is given locally near $z_0$ by $f = t^m$, and the function $g$ is given locally near $x_0$ by the function $g = u^n$. Then in a neighbourhood of $(z_0,x_0)$, the equation $f(z) = g(x)$ can be written as $t^m = u^n.$ The curve cut out by this equation is singular at $(0,0)$ precisely when both $m$ and $n$ are $> 1$.
Conclusion: The curve $Y$ will be singular if there is a point in $N$ that is ramified for both $f$ and $g$, otherwise it will be smooth (i.e. a Riemann surface).