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Given any triple of positive integers $(g',g,d)$ with $2g'-2\geq d(2g-2)$.

Does there always exist curves $C_{g'},C_g$ of genus $g',g$ with a degree $d$ morphism $f\colon C_{g'}\to C_g$?

If we fix $C_g$ a curve of genus $g$, can we always find a a curve $C_{g'}$ with genus $g'$ and a degree $d$ morphism $f\colon C_{g'}\to C_g$?

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Yes. Let me assume for simplicity $g\geq 2$ — the cases $g=0,1$ can be treated in the same way with slight modifications. Put $r=2g'-2-d(2g-2)$, and choose $r+1$ distinct points $p;p_1,\ldots ,p_r$ on $C_g$. The group $\pi:=\pi _1(C_g\smallsetminus \{p_1,\ldots ,p_r\},p )$ is generated by $2g+r$ elements $a_1,\ldots ,a_g;b_1,\ldots ,b_g;c_1,\ldots ,c_r$, with the relation $c_1\ldots c_r\prod (a_i,b_i)=1$. Let $\sigma ,\tau $ be two permutations which generate $\mathfrak{S}_d$. There is a surjective homomorphism $\pi \rightarrow \mathfrak{S}_d$ which maps each $c_i$ to the transposition $(1,2)$, $a_1$ to $\sigma $, $a_2$ to $\tau $, and all others $a_i$, $b_j$ to $1$ (note that $r$ is even). It corresponds to a connected covering $\tilde{C}\rightarrow C_g $ with a simple ramification point above each $p_i$, and étale elsewhere. By the Riemann-Hurwitz formula $g(\tilde{C} )=g'$.

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  • $\begingroup$ If we choose points $p_i$ on $C_g$ with prescribed ramification index $e_i$, such that $\sum (e_i-1)=2g-2-d(2g'-2)$. What will be the number of isomorphic classes of branched covering $C_g'\to C_g$? $\endgroup$
    – user39380
    Commented Apr 28, 2015 at 1:56
  • $\begingroup$ No, it is much more complicated. Given such a choice, you must still choose a homomorphism $\pi _1\rightarrow \mathfrak{S}_d$ with certain properties, modulo conjugation. This is a difficult and interesting question -- google "Hurwitz numbers". $\endgroup$
    – abx
    Commented Apr 28, 2015 at 3:48
  • $\begingroup$ Sorry, does $\pi_1\to S_d$ gives a degree $d!$ covering? If we choose a quotient of $S_d$ of order $d$, we get a $d$ sheet covering. How to read the ramification $\endgroup$
    – user39380
    Commented Apr 28, 2015 at 12:21
  • $\begingroup$ You better see this homomorphism as an action of $\pi _1$ on a set $E$ with $d$ elements. The fiber at $p_i$ is naturally identified with the quotient of $E$ by the action of $c_i\in \pi _1$ (the class of a small loop around $p_i$). For instance, you have only one simple ramification point iff the image of $c_i$ in $\mathfrak{S}_d$ is a transposition. $\endgroup$
    – abx
    Commented Apr 28, 2015 at 15:53

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