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Consider that for a meromorphic function $f$ on a Riemann surface, we have an associated holomorphic map $F: X \to \mathbb{C} \cup \{ \infty \}$ to the Riemann sphere defined by $$F(z) = \begin{cases} f(z), & z \ \text{not a pole of $f$} \\ \infty, & \text{otherwise}. \end{cases}$$ I want to show that the map $\phi : X \to \mathbb{C}\text{P}^1$ defined by $z \longmapsto [1 : f(z)]$ can be realised as the above map $F$.

It is clear that $\phi(z) = [1:f(z)]$ is defined for $z$ not a pole of $f$. But I don't see how $[1 : f(z)] \sim f(z)$ and how we have a notion of $\infty$ in projective space.

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  • $\begingroup$ @hwong557 Could you elaborate a little more? $\endgroup$
    – user319128
    Feb 12 '17 at 4:30
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The point $[0:1]$ can be realized as $\infty$. Perhaps here is some motivation. Suppose we have two polynomials $p(z)$ and $q(z)$ on $\mathbb C$. Then clearly $p(z)/q(z)$ is a meromorphic function on $\mathbb C$. As you said, we can realize this as a map to $\mathbb P^1$ by sending $z$ to $[1:p(z)/q(z)]$, which is the same as $[q(z):p(z)]$. Notice $p/q$ has a pole at $z_0$ precisely if $q(z_0)=0$. This corresponds to the point $[0:p(z_0)] = [0:1]$, which is thought of as $\infty$.

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