proof of that theorem, that some of these factors are (isogenous to) the elliptic curves $$y^{2}=x(x-1)(x-\lambda)$$ $$y^{2}=x(x-\mu_{j,1})(x-\mu_{j,2}), \quad j=1,\ldots,s-2,$$ $$y^{2}=(x-1)(x-\lambda)(x-\mu_{j,2}), \quad j=1,\ldots,s-2.$$

The choice we have done for the tuple $(\lambda,\mu_{1,1},\mu_{1,2},\mu_{2,1},\mu_{2,2},\ldots,\mu_{s-2,1},\mu_{s-2,2}) \in \Delta_{2s-3}$ ensures that they are isomorphic to the elliptic curves $$y^{2}=x(x-1)(x-\rho)$$ where $\rho \in \{\lambda_{1},\ldots,\lambda_{r}\}$.

The case $r \geq 4$ even can be worked similarly, but in this case we add an extra elliptic curve to the $r$ given ones in order to obtain the result as a consequence of the odd situation.

An irreducible fiber product of elliptic curves: Theorem \[construccion2\] {#Sec:otraconstruccion}
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In the above main construction, the fiber product turned-out to be reducible; it has two isomorphic irreducible components. In this section we describe a similar construction, but the fiber product we obtain is irreducible. Unfortunately, the genus is bigger than the previous one.

The fiber product construction
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Let $(\lambda_{1},\ldots,\lambda_{r}) \in \Delta_{r}$, $E_{\lambda_{j}}: y_{j}^{2}=x_{j}(x_{j}-1)(x_{j}-\lambda_{j})$ and $\pi_{j}:E_{\lambda_{j}} \to \widehat{\mathbb C}$ defined by $\pi_{j}(x_{j},y_{j})=x_{j}$. The locus of branch values of $\pi_{j}$ is the set $\{\infty,0,1,\lambda_{j}\}$. Let $X$ be the fiber product of these $r$ pairs $(E_{\lambda_{1}},\pi_{1}),\ldots,(E_{\lambda_{r}},\pi_{r})$, that is, $$X=\left\{(x,y_{1},\ldots,y_{r}): y_{j}^{2}=x(x-1)(x-\lambda_{j}), \; j=1,\ldots,r \right\}.$$

The curve $X$ is irreducible but it has singular points at those points 