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Is it true that if $X$ and $Y$ are two irreducible algebraic curves over $\mathbb{C}$ and $f:X\dashrightarrow Y$ is a rational map that is generically n:1 for some $n\in \mathbb{Z}^{+}$ then the geometric genus of $Y$ is less than or equal to the geometric genus of $X$?

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    $\begingroup$ Replace $X,Y$ by their smooth projective models and use Riemann-Hurwitz to show the desired inequality. $\endgroup$
    – Mohan
    Commented Apr 17, 2017 at 17:54

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Mohan' answer in the comments is correct: this is more a comment to his answer.

If you are only interested in the inequality you do not need Riemann-Hurwitz formula as I will explain in the following. The advantage is that it generalizes to higher dimension.

Replace first, as Mohan suggests, the irreducible curves with their normalizations. Then you can pull-back holomorphic 1-forms. Since a finite map is a local diffeomorphism on a (dense) open set, the pull-back of a 1-form not identically zero is not the zero form. In other word the pull-back is an injective linear map among the respective vector space of holomorphic 1-forms whose dimensions are by definition the respective geometric genera. This proves the inequality.

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