$JS$ is isogenous to $E_{1} \times E_{2}$.

The automorphism $A_{2}\circ A_{1}(x,y_{1},y_{2})=(x,-y_{1},-y_{2})$ is the hyperelliptic one. Let us consider a two-fold branched cover $P:C \to \widehat{\mathbb C}$ with deck group $\langle A_{2}\circ A_{1} \rangle$. We may assume that the involution $A_{1}$ descent to the involution $T(x)=-x$ (this because all involutions are conjugated in the Möbius group). So the branch values of $P$ may be assume to be $\pm 1$, $\pm \rho_{1}$ and $\pm \rho_{2}$. In this way, $C$ can be described by the hyperelliptic curve $$y^{2}=(x^{2}-1)(x^{2}-\rho_{1}^{2})(x^{2}-\rho_{2}^{2}).$$

Let us consider the two-fold branched cover $Q(x)=x^{2}$ whose deck group is $\langle T \rangle$. The branch values of $Q$ are $1$, $\rho_{1}^{2}$ and $\rho_{2}^{2}$. The images of $\infty$ and $\pm 1$ are $\infty$ and $1$. Let us consider the Möbius transformation $$L(x)=\frac{(1-\rho_{1}^{2})(x-\rho_{2}^{2})}{(1-\rho_{2}^{2})(x-\rho_{1}^{2})}.$$

Then $L(1)=1$, $L(\rho_{1}^{2})=\infty$, $L(\rho_{2}^{2})=0$, $L(\infty)=(1-\rho_{1}^{2})/(1-\rho_{2}^{2})$ and $L(0)=\rho_{2}^{2}L(\infty)/\rho_{1}^{2}$. By making $L(\infty)=\lambda_{1}$ and $L(0)=\lambda_{2}$, we obtain that $$\rho_{1}^{2}=\frac{\lambda_{1}-1}{\lambda_{2}-1}, \quad \rho_{2}^{2}=\frac{\lambda_{2}}{\lambda_{1}} \rho_{1}^{2}$$

Proof of Theorem \[construccion\]
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Let us consider the following $s$ elliptic curves $$\begin{array}{ll}
E_{1}: & y^{2}=x(x-1)(x-\lambda)\\
E_{2}: & y^{2}=(x-1)(x-\lambda)(x-\mu_{1,1})(x-\mu_{1,2})\\
E_{3}: & y^{2}=(x-\mu_{1,1})(x-\mu_{1,2})(x-\mu_{2,1})(x-\mu_{2,2})\\
E_{4}: & y^{2}=(x-\mu_{2,1})(x-\mu_{2,2})(x-\mu_{3,1})(x-\mu_{3,2})\\
&\vdots\\
E_{s-1}: & y^{2}=(x-\mu_{r-3,1})(x-\mu_{s-3,2})(x-\mu_{s-2,1})(x-\mu_{s-2,2})\\
E_{s}: & y^{2}=x(x-\mu_{s-2,1})(x-\mu_{s-2,2})
\end{array}$$

If we consider the degree two maps $\pi_{j}:E_{j} \to \widehat{\mathbb C}$ defined as $\pi_{j}(x,y)=x$, then we may perform the 