It is well known that the Abel-Jacobi map restricted to $\text{Eff}\_g(C)$ surjects onto the Jacobian $\text{Jac}(C)$, since every divisor of degree $g$ is effective. 


Is there an analogous statement for Prym varieties? That is, given an unramified double cover $\widetilde C\to C$ with involution $\tau$, consider the map $f:\text{Eff}\_d(\widetilde{C})\to\text{Prym}(\widetilde{C}/C)$ given by $f(D)=D-\tau(D)$. Is $f$ surjective if, for instance, $d=g-1$?



Surjectivity of the Abel-Prym map
First of all, note that your definition is not correct: when $d$ is odd, the image of your map does not land in the Prym variety -- you have to add a constant term. When this is done, the answer is yes, for the following reason. Let $X$ be the image of $\tilde{C} $ in $P:=\operatorname{Prym}(\tilde{C}/C ) $. Let me put $h:=g-1=\dim P$. What you want to prove is that the addition map $X^{h}\rightarrow P$ is surjective, that is, of degree $>0$. Now this degree is computed by the Pontryagin product $[X]^{\*h}$, where $[X]$ is the class of $X$ in $H^{2h-2}(P,\mathbb{Z})$. We know that this class is $2\dfrac{\theta ^{h-1}}{(h-1)!} $, where $\theta $ is the class of the principal polarization. 


So we just have to prove that $\theta ^{\*h}\in H^{2h}(P,\mathbb{Z})$ is nonzero. This is true for any principally polarized abelian variety $(P,\theta )$ of dimension $h$: it suffices to prove it for a Jacobian $J\Gamma $, and this amounts to say that the Abel-Jacobi map $\Gamma ^h\rightarrow J\Gamma $ is surjective, as you recall in your post.


