Let $X$ be a hyperelliptic curve and let $i:X\to X$ denote the hyperelliptic involution. Once we fix a point $x\_0\in X$ we get the Abel-Jacobi map $AJ:X\to J$ where $J$ denotes the Jacobian variety. Now the Jacobian is also equipped with an involution, namely $x\mapsto x^{-1}$. Is it possible to choose the base point $x\_0$ in such a way that the restriction of the involution on the Jacobian is the involution on $X$. 



Involution on Hyperelliptic curves and their Jacobians
Yes, pick $x\_0$ to be a Weierstrass point, i.e. a fixed point of the hyperelliptic involution. 


Let $\sigma$ denote the hyperelliptic involution on $X$. Under the Abel-Jacobi map we have $x \mapsto [x-x\_0]$ and $\sigma(x) \mapsto [\sigma(x) - x\_0]$. Now $[x + \sigma(x) - 2x\_0]$ is the divisor of a function, since it is the pullback under the hyperelliptic map of a degree zero divisor on $\mathbf P^1$. Hence $AJ(x)$ and $AJ(\sigma(x))$ are inverses.


