A curve in an abelian surface and its image in the Kummer surface This is related to a question that I already asked Curve through the 16 singular points of a Kummer surface. I am new to Abelian varieties and I am trying to understand more about them. 
Let $X=J(C)$ be the Jacobian of a genus 2 curve over $\mathbb{C}$, so $X$ is an abelian surface. We can choose an embedding of $C$ in $X$ such that the hyperelliptic involution on $C$ extends to the involution on $X$. Consider $L=\mathcal{O}(nC)$, where $n$ is an even positive integer. Then $L$ is a totally symmetric line bundle. So $L=f^*L'$ where $f:J(C) \longrightarrow K(C)$ is the quotient map from the Jacobian to the associated Kummer surface, and $L'$ is an ample line bundle on $K(C)$. 
Consider a general smooth curve $C'\in |L|$ through all sixteen 2-torsion points. Let $C''$ be its image in $K(C)$, this passes through all sixteen singular points of $K(C)$.
a) For a general smooth $C'$ as above, will the restriction  of $f$ from $C' \longrightarrow C''$ be a double cover, or an isomorphism or what?
b) For a general smooth $C'$ as above, is the image $C''$ smooth?
c) When is such a $C'$ isomorphic to its image $C''$?
Thanks in advance for the help!
 A: Let $b:\hat{X}\rightarrow X$ the blowing up of the 16 2-torsion points, and $E_1,\ldots ,E_{16}$ the exceptional  $(-1)$ curves on $\hat{X}$. The involution $x\mapsto -x$ lifts to an involution $\sigma $ of $\hat{X}$. 
You are looking at the linear system $|nb^*C|-\sum E_i$. The involution $\sigma $ acts on $H=H^0(\hat{X}, \mathcal{O}(nb^*C-\sum E_i))$, which splits as $H^+\oplus H^-$. The fixed curves of the system are those defined by elements of $H^+$ or $H^-$. Using for instance  the holomorphic Lefschetz formula, one sees that these two subspaces are nonzero for $n\gg 0$. Thus a general $C'$ in the system is not  fixed by $\sigma $, and therefore maps birationally to the curve $C''\subset K(C)$. However that map is not an isomorphism :  there are pairs of points of $C'$ which map to the same point of $C''$, namely those which lie also in $\sigma (C')$. Since $\sigma $ acts trivially on $H^2(\hat{X})$, we have $C'\cdot \sigma (C')= (C')^2=2n^2-16$, so there are plenty of such points. From general principles it should be easy to show that for $C'$ general the intersection consists of $2n^2-16$ distinct points, so that $C''$ has $2n^2-16$ ordinary double points and no other singularities, and $C'\rightarrow C''$ is its normalization.
