Let $f:C\longrightarrow C'$ be a finite degree 2 morphism of smooth projective curves. If the gonality$(C')=k$, then we can say that the gonality$(C)\leq 2k$. Under what conditions is the gonality exactly $2k$? Can we impose some conditions on the genus of curves for example such that this happens?
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3$\begingroup$ C' has a map g to P^1 of degree 2k given by composition. If it has another map h to P^1 of smaller degree, you should be able to show (maybe with some extra assumptions) that (g,h): C' -> P^1 x P^1 is birational onto its image. But the image is curve of bidegree at most (2k,2k) so its geometric genus is on order 4k^2 at most. If the genus of C is massively higher than this, you should get a contradiction. So that's one situation where you might know gon(C') = 2k. $\endgroup$– JSENov 3, 2015 at 15:42
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1$\begingroup$ Jordan's suggestion will work in certain circumstances but not all. If $C$ is hyperelliptic (so $C'$ is also, this can be easily arranged), then both will have gonality $2$ and you can make the genus of $C,C'$ pretty much anything you want. $\endgroup$– Felipe VolochNov 3, 2015 at 15:48
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$\begingroup$ @FelipeVoloch You can't quite make the genus of $C$ and $C'$ arbitrary (at least not at the same time). If $C$ is hyperelliptic and $C\to C'$ is a double cover, then $g(C') = g(C)/2$ if $g(C)$ is even, and $g(C')=(g(C)\pm 1)/2$ if $g(C)$ is odd. $\endgroup$– Dan PetersenNov 3, 2015 at 21:17
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$\begingroup$ @DanPetersen OP didn't say anything about covers being unramified so $g \ge 2g' - 1$ is the only restriction. $\endgroup$– Felipe VolochNov 3, 2015 at 22:37
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$\begingroup$ @FelipeVoloch I didn't assume the cover is unramified either. The covering map gives an extra involution of $C$, which must come from an involution of $\mathbf P^1$ permuting the branch points of the hyperelliptic map. It follows that $C$ has affine equation $y^2=f(x^2)$ and $C'$ has equation $y^2=f(x)$ or $y^2=xf(x)$. $\endgroup$– Dan PetersenNov 4, 2015 at 7:38
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