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Let $Y_0$ be a genus two projective smooth complex curve, let $Y_1$ be an étale cover of degree 2 of $Y_0$, and let $\sigma$ be the involution of $Y_1$ over $Y_0$. If $J_1$ is the jacobian of $Y_1$, we let $P$ be the image $(1-\sigma)(J_1)$. This Prym variety $P$ is an elliptic curve. Look at the map $f$ induced by $1-\sigma$ from $J_1[2]$ to $P[2]$.

Question: Can it happen that $f$ is surjective?

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It is always surjective (regardless of the genus of $Y_0$). The point is that the kernel $K$ of $1-\sigma :J_1\rightarrow P$ is connected — it is equal to the image of the pull back map $J_0\rightarrow J_1$. Now apply the snake lemma to multiplication by 2 in the exact sequence $0\rightarrow K\rightarrow J_1\xrightarrow{\ 1-\sigma \ }P\rightarrow 0$.

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  • $\begingroup$ The kernel of $1 - \sigma$ is not connected for an \'{e}tale cover it is only connected when $\sigma$ has fixed points. For $\sigma$ with no fixed points the kernel has two connected components: the image of $J_{0} \to J_{1}$ and a translate of that image by a point of order $2$ on $J_{1}$. $\endgroup$ Commented Feb 25, 2022 at 13:12
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    $\begingroup$ @Tony Pantev: I don't think so. If $L$ is a line bundle on $Y_1$ invariant under $\sigma $, and $A$ is a very ample line bundle on $Y_0$, $\sigma $ acting on the linear system $\lvert L\otimes \pi ^*A\rvert\cong \mathbb{P}^m$ has some fixed points; such a fixed point is a $\sigma $-invariant effective divisor, hence of the form $\pi ^*D$, and $L=\pi ^*(A^{-1}(D))$. $\endgroup$
    – abx
    Commented Feb 25, 2022 at 14:38
  • $\begingroup$ Ah, sorry - my mistake! The statement I was thinking about is that for an \'{e}tale double cover the kernel of $1 + \sigma$ is disconnected and consists of $P$ and a translate of $P$ by a point of order $2$. $\endgroup$ Commented Feb 25, 2022 at 16:11
  • $\begingroup$ So, yes, you are absolutely right - the kernel K is connected and the snake lemma argument works. $\endgroup$ Commented Feb 25, 2022 at 17:23

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