The fibres of lambda are exactly the Gamma(2)-orbits

Put $L_j=e_j(\tau )$ and $L_j^{\mathrm{\prime }}=e_j({\tau }^{\mathrm{\prime }})$. If $\lambda (\tau )=\lambda ({\tau }^{\mathrm{\prime }})$ then with $u:=L_3-L_2$, $v:=L_1-L_2$, $u^{\mathrm{\prime }}:=L_3^{\mathrm{\prime }}-L_2^{\mathrm{\prime }}$, $v^{\mathrm{\prime }}:=L_1^{\mathrm{\prime }}-L_2^{\mathrm{\prime }}$ we have $u\mathrm{/}v=u^{\mathrm{\prime }}\mathrm{/}{v}^{\mathrm{\prime }}$, so $u^{\mathrm{\prime }}v=u{v}^{\mathrm{\prime }}$. Hence $(L_3^{\mathrm{\prime }}-L_2^{\mathrm{\prime }})(L_1-L_2)=(L_3-L_2)(L_1^{\mathrm{\prime }}-L_2^{\mathrm{\prime }})$, i.e. $(L_3^{\mathrm{\prime }}-L_2^{\mathrm{\prime }})(L_1-L_2)-(L_3-L_2)(L_1^{\mathrm{\prime }}-L_2^{\mathrm{\prime }})=0$, a determinant condition; the affine map $\mathrm{\Phi }(x)=\alpha x+\beta$ with $\alpha :=\frac{L_2^{\mathrm{\prime }}-L_1^{\mathrm{\prime }}}{L_2-L_1}$ and $\beta :=L_1^{\mathrm{\prime }}-\alpha L_1$ satisfies $\mathrm{\Phi }(L_1)=L_1^{\mathrm{\prime }}$ and $\mathrm{\Phi }(L_2)=L_2^{\mathrm{\prime }}$, and the displayed identity says exactly $\mathrm{\Phi }(L_3)=L_3^{\mathrm{\prime }}$. Since $\sum L_j=\sum L_j^{\mathrm{\prime }}=0$ we get $\beta =\frac{1}{3}(\sum L_j^{\mathrm{\prime }})-\alpha \frac{1}{3}(\sum L_j)=0$, so $L_j^{\mathrm{\prime }}=\alpha L_j$ for $j=1,2,3$ with $\alpha \mathrm{\ne }0$.

Choose a square root $a$ of ${\alpha }^{-1}$ and put ${\mathrm{\Lambda }}^{\ast }:=a{\mathrm{\Lambda }}_{\tau }$. By [F2], $G_4({\mathrm{\Lambda }}^{\ast })=a^{-4}{G}_4({\mathrm{\Lambda }}_{\tau })={\alpha }^2{G}_4({\mathrm{\Lambda }}_{\tau })$ and $G_6({\mathrm{\Lambda }}^{\ast })=a^{-6}{G}_6({\mathrm{\Lambda }}_{\tau })={\alpha }^3{G}_6({\mathrm{\Lambda }}_{\tau })$, hence $g_2({\mathrm{\Lambda }}^{\ast })={\alpha }^2{g}_2(\tau )$ and $g_3({\mathrm{\Lambda }}^{\ast })={\alpha }^3{g}_3(\tau )$. On the other hand $e_j({\tau }^{\mathrm{\prime }})=\alpha e_j(\tau )$ by 1.1 and the invariants are the elementary symmetric functions of the three branch values [F2], so $g_2({\tau }^{\mathrm{\prime }})={\alpha }^2{g}_2(\tau )$ and $g_3({\tau }^{\mathrm{\prime }})={\alpha }^3{g}_3(\tau )$ as well; therefore ${\mathrm{\Lambda }}^{\ast }$ and ${\mathrm{\Lambda }}_{{\tau }^{\mathrm{\prime }}}$ have the same invariants $g_2,g_3$.

By 2.1 the lattices ${\mathrm{\Lambda }}^{\ast }$ and ${\mathrm{\Lambda }}_{{\tau }^{\mathrm{\prime }}}$ have the same invariants, so their Weierstrass cubics are identical. Their biholomorphisms [F3] to this cubic induce a biholomorphism of tori fixing the origin and matching the three labelled half-periods: the branch values for ${\mathrm{\Lambda }}^{\ast }=a{\mathrm{\Lambda }}_{\tau }$ are $a^{-2}{L}_j=\alpha L_j=L_j^{\mathrm{\prime }}$. Lift this map and its inverse to based maps of $\mathbb{C}$ using the holomorphic lattice coverings, the lifting criterion and simple connectedness of $\mathbb{C}$; uniqueness of based lifts makes the lifts inverse biholomorphisms. The entire-biholomorphism theorem gives a lift $z\mapsto bz$, $b\mathrm{\ne }0$. Therefore multiplication by $A:=ba$ carries ${\mathrm{\Lambda }}_{\tau }$ onto ${\mathrm{\Lambda }}_{{\tau }^{\mathrm{\prime }}}$ and matches the labelled half-periods modulo these lattices. This labelled homothety, rather than the false Laurent recursion previously recorded, is sufficient for the final congruence calculation.

Since ${\mathrm{\Lambda }}_{{\tau }^{\mathrm{\prime }}}=A{\mathrm{\Lambda }}_{\tau }$, the numbers $A$ and $A\tau$ form a positively oriented basis of ${\mathrm{\Lambda }}_{{\tau }^{\mathrm{\prime }}}$ (multiplication by $A$ preserves orientation), so $A=p+q{\tau }^{\mathrm{\prime }}$ and $A\tau =r+s{\tau }^{\mathrm{\prime }}$ for integers $p,q,r,s$ forming a matrix ${\gamma }_0=\left(\begin{array}{cc}p& r\\ q& s\end{array}\right)\in S{L}_2(\mathbb{Z})$. The labelled homothety from 3.1 gives $A{h}_j(\tau )\equiv h_j(\tau \mathrm{\prime })(\mathrm{m}\mathrm{o}\mathrm{d}{\mathrm{\Lambda }}_{\tau \mathrm{\prime }})$ for $j=1,2,3$. Taking $j=1$ gives $p+q{\tau }^{\mathrm{\prime }}=A\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2{\mathrm{\Lambda }}_{{\tau }^{\mathrm{\prime }}})$, so $p$ is odd and $q$ even; taking $j=2$ gives $A\tau =r+s{\tau }^{\mathrm{\prime }}\equiv {\tau }^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}2{\mathrm{\Lambda }}_{{\tau }^{\mathrm{\prime }}})$, so $r$ is even and $s$ odd; hence ${\gamma }_0\equiv I(\mathrm{m}\mathrm{o}\mathrm{d}2)$ [F5]. Finally $A\tau =r+s{\tau }^{\mathrm{\prime }}$ and $A=p+q{\tau }^{\mathrm{\prime }}$ give ${\tau }^{\mathrm{\prime }}=\frac{p\tau -r}{s-q\tau }$, that is ${\tau }^{\mathrm{\prime }}=\gamma \cdot \tau$ for $\gamma =\left(\begin{array}{cc}p& -r\\ -q& s\end{array}\right)$, which has determinant $ps-qr=1$ and entries congruent to $I$ modulo $2$, so $\gamma \in \mathrm{\Gamma }(2)$ [F5]. Conversely $\lambda$ is $\mathrm{\Gamma }(2)$-invariant [F5], so the fibres of $\lambda$ are exactly the $\mathrm{\Gamma }(2)$-orbits.