# Is the class of $k$-gonal curves dominant

Before I start, let me make a note on terminology. Curves are always smooth projective connected curves over an algebraically closed field of characteristic zero.

Let $\mathcal C$ be a class of curves. We say that $\mathcal C$ is dominant if, for all curves $X$, there exists a curve $Y$ in $\mathcal C$ and a finite morphism $Y\to X$.

Bogomolov and Tschinkel proved that the class of hyperelliptic curves and their unramified covers is dominant. Manin proved that the class of modular curves $X(n)$ and their unramified covers is dominant. Both proofs rely on Abhyankar's Lemma.

Let $k\geq 2$ be an integer. Let $\mathcal C_k$ be the class of $k$-gonal morphisms, i.e., the class of curves for which the gonality equals $k$.

Q1. Is $\mathcal{C}_2$ dominant?

Q2. Is $\mathcal{C}_k$ dominant?

Q3. Is $\cup_{2 \leq j \leq k} \mathcal{C}_j$ dominant if $k>>0$?

Let me repeat this in words. Let $X$ be a curve. Does there exist a $k$-gonal curve $Y$ and a finite morphism $Y\to X$?

I'm mainly interested in the case $k=2$. In this case, it suffices to answer the following question.

Q1b. Let $X$ be a curve. Does $\mathbf{P}^1$ admit a closed immersion into the symmetric product $X^{(2)}$?

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The points of $X^{(2)}$ correspond to divisors of degree 2; two points of $X^{(2)}$ are in the image of a morphism $\mathbb P^1 \to X^{(2)}$ if and only if the two corresponding divisors are linearly equivalent. Hence there is a non-trivial morphism $\mathbb P^1 \to X^{(2)}$ if and only if $X$ is hyperelliptic.
More generally, if $Y \to X$ is non-constant morphism and $Y$ is $k$-gonal, then $X$ is also $k$-gonal.
Thank you for your answer. I didn't realize that the existence of a non-constant morphism from $\mathbf{P}^1$ to $X^{(k)}$ implied the existence of a $g^1_k$ on $X$. – Ariyan Javanpeykar Jul 19 '12 at 13:42
The final assertion in the answer should be that if $Y$ is $k$-gonal then $X$ is $\ell$-gonal for some $\ell\le k$. – Michael Zieve May 12 '14 at 19:23