[proofplan] We first prove existence by choosing, among all transforms of a given point $z \in \mathbb H$, one with maximal imaginary part. This is possible because the quantities $cz+d$ form a discrete subset of $\mathbb C$ as $(c,d)$ ranges over primitive integer pairs. We then translate horizontally by a power of $T$ to enter the standard vertical strip, and maximality forces the point to lie outside the unit disk. For uniqueness in the interior, the same imaginary-part formula shows that an equivalent interior point cannot come from a matrix with nonzero lower-left entry; the remaining translation must be trivial. Finally, the equality cases in the inequalities defining $\mathcal F$ give precisely the vertical and circular boundary pairings. [/proofplan]

[step:Compute the imaginary part of a modular transform]Let \begin{align*} \gamma := \begin{pmatrix}a&b\\c&d\end{pmatrix} \in SL_2(\mathbb Z) \end{align*} and let $z=x+iy \in \mathbb H$, where $x:=\operatorname{Re}(z)$ and $y:=\operatorname{Im}(z)>0$. Since $ad-bc=1$, we have \begin{align*} \operatorname{Im}(\gamma z) &= \operatorname{Im}\left(\frac{az+b}{cz+d}\right) \\ &= \operatorname{Im}\left(\frac{(az+b)(c\overline z+d)}{|cz+d|^2}\right) \\ &= \frac{\operatorname{Im}(ac|z|^2+ad z+bc\overline z+bd)}{|cz+d|^2} \\ &= \frac{(ad-bc)y}{|cz+d|^2} \\ &= \frac{y}{|cz+d|^2}. \end{align*} Thus maximizing $\operatorname{Im}(\gamma z)$ is equivalent to minimizing $|cz+d|$ over all bottom rows $(c,d)$ of matrices in $SL_2(\mathbb Z)$.[/step]

[guided]The key quantity is the denominator $cz+d$. Let \begin{align*} \gamma := \begin{pmatrix}a&b\\c&d\end{pmatrix} \in SL_2(\mathbb Z), \end{align*} so $a,b,c,d \in \mathbb Z$ and $ad-bc=1$. For $z=x+iy \in \mathbb H$, with $y>0$, compute the imaginary part of the fractional linear transformation: \begin{align*} \gamma z = \frac{az+b}{cz+d}. \end{align*} Multiplying numerator and denominator by $c\overline z+d$ gives \begin{align*} \gamma z = \frac{(az+b)(c\overline z+d)}{|cz+d|^2} = \frac{ac|z|^2+ad z+bc\overline z+bd}{|cz+d|^2}. \end{align*} The terms $ac|z|^2$ and $bd$ are real. Since $\operatorname{Im}(z)=y$ and $\operatorname{Im}(\overline z)=-y$, the imaginary part of the numerator is \begin{align*} ad\,y+bc(-y)=(ad-bc)y=y. \end{align*} Therefore \begin{align*} \operatorname{Im}(\gamma z)=\frac{y}{|cz+d|^2}. \end{align*} This formula is the engine of the proof: making the transformed point as high as possible in $\mathbb H$ is the same as making $|cz+d|$ as small as possible.[/guided]

[step:Choose a transform with maximal imaginary part]Let \begin{align*} \Lambda_z := \{cz+d : c,d \in \mathbb Z\} \subset \mathbb C. \end{align*} This is the lattice generated by $z$ and $1$. Since $\operatorname{Im}(z)>0$, the real-[linear map](/page/Linear%20Map) \begin{align*} A_z: \mathbb R^2 &\to \mathbb C \\ (u,v) &\mapsto uz+v \end{align*} is an isomorphism of real vector spaces, so $A_z(\mathbb Z^2)=\Lambda_z$ is a lattice and therefore a discrete subset of $\mathbb C$. In particular, $0$ is isolated in $\Lambda_z$ and every bounded disk in $\mathbb C$ contains only finitely many points of $\Lambda_z$: otherwise the inverse image under $A_z^{-1}$ would give infinitely many points of $\mathbb Z^2$ in a bounded subset of $\mathbb R^2$, which is impossible. Hence the set \begin{align*} \{|cz+d| : (c,d)\in \mathbb Z^2,\ (c,d)\neq (0,0)\} \end{align*} has a positive minimum, and that minimum is attained. Choose a pair $(c_0,d_0)\in \mathbb Z^2$ attaining this minimum. This pair is primitive: if $(c_0,d_0)=k(c_1,d_1)$ with $c_1,d_1,k\in\mathbb Z$ and $|k|>1$, then \begin{align*} |c_1z+d_1|=\frac{1}{|k|}|c_0z+d_0|<|c_0z+d_0|, \end{align*} contradicting minimality. Since $\gcd(c_0,d_0)=1$, [Bezout's identity](/page/Bezout%20Identity) gives $a_0,b_0\in \mathbb Z$ such that \begin{align*} a_0d_0-b_0c_0=1. \end{align*} Define \begin{align*} \gamma_0 := \begin{pmatrix}a_0&b_0\\c_0&d_0\end{pmatrix}\in SL_2(\mathbb Z). \end{align*} By the imaginary-part formula, $\operatorname{Im}(\gamma_0 z)$ is maximal among all $\operatorname{Im}(\gamma z)$ with $\gamma\in SL_2(\mathbb Z)$.[/step]

[guided]We need a transform of $z$ that is as high as possible in the upper half-plane. By the formula already proved, this means we need to minimize $|cz+d|$ among admissible lower rows of matrices in $SL_2(\mathbb Z)$. Define \begin{align*} \Lambda_z := \{cz+d : c,d \in \mathbb Z\} \subset \mathbb C. \end{align*} This is a lattice because it is the image of $\mathbb Z^2$ under the real-linear map \begin{align*} A_z: \mathbb R^2 &\to \mathbb C \\ (u,v) &\mapsto uz+v. \end{align*} The map $A_z$ is invertible over $\mathbb R$: if $uz+v=0$ with $u,v\in\mathbb R$, then taking imaginary parts gives $u\operatorname{Im}(z)=0$, hence $u=0$, and then $v=0$. Therefore $A_z(\mathbb Z^2)$ is a lattice and hence a discrete subset of $\mathbb C$. More explicitly, $0$ is isolated in $\Lambda_z$, and every bounded disk in $\mathbb C$ contains only finitely many points of $\Lambda_z$, because applying $A_z^{-1}$ would otherwise put infinitely many points of $\mathbb Z^2$ in a bounded subset of $\mathbb R^2$. Because of this local finiteness, the nonzero lengths \begin{align*} \{|cz+d| : (c,d)\in \mathbb Z^2,\ (c,d)\neq (0,0)\} \end{align*} have a positive minimum, and the minimum is attained. Choose a pair $(c_0,d_0)$ attaining this minimum. This pair is primitive. Indeed, if $(c_0,d_0)=k(c_1,d_1)$ with $c_1,d_1,k\in\mathbb Z$ and $|k|>1$, then \begin{align*} |c_1z+d_1|=\frac{1}{|k|}|c_0z+d_0|<|c_0z+d_0|, \end{align*} which contradicts the choice of $(c_0,d_0)$ as a minimizer. Since $\gcd(c_0,d_0)=1$, [Bezout's identity](/page/Bezout%20Identity) gives integers $a_0,b_0\in\mathbb Z$ such that \begin{align*} a_0d_0-b_0c_0=1. \end{align*} Thus \begin{align*} \gamma_0 := \begin{pmatrix}a_0&b_0\\c_0&d_0\end{pmatrix} \end{align*} lies in $SL_2(\mathbb Z)$. For every $\gamma\in SL_2(\mathbb Z)$, its lower row $(c,d)$ is a nonzero integer pair, so \begin{align*} |c_0z+d_0|\le |cz+d|. \end{align*} Using \begin{align*} \operatorname{Im}(\gamma z)=\frac{\operatorname{Im}(z)}{|cz+d|^2}, \end{align*} we conclude that $\operatorname{Im}(\gamma_0 z)$ is maximal among all modular transforms of $z$.[/guided]

[step:Translate the maximal transform into the standard strip] Put $w_0:=\gamma_0 z$. Choose $n\in\mathbb Z$ such that \begin{align*} -\frac{1}{2}\le \operatorname{Re}(w_0)+n\le \frac{1}{2}. \end{align*} Define \begin{align*} w:=T^n w_0=w_0+n. \end{align*} Then $w$ is $SL_2(\mathbb Z)$-equivalent to $z$ and satisfies \begin{align*} -\frac{1}{2}\le \operatorname{Re}(w)\le \frac{1}{2}. \end{align*} Since $T^n$ has lower row $(0,1)$, the operation $w_0\mapsto w_0+n$ preserves imaginary part: \begin{align*} \operatorname{Im}(w)=\operatorname{Im}(w_0). \end{align*} Hence $w$ also has maximal imaginary part among all $SL_2(\mathbb Z)$-transforms of $z$. [/step]

[step:Use maximality to force the unit circle inequality]Suppose, for contradiction, that $|w|<1$. Since \begin{align*} S w=-\frac{1}{w}, \end{align*} we have \begin{align*} \operatorname{Im}(Sw) = \operatorname{Im}\left(-\frac{1}{w}\right) = \frac{\operatorname{Im}(w)}{|w|^2} > \operatorname{Im}(w). \end{align*} This contradicts the maximality of $\operatorname{Im}(w)$ among all $SL_2(\mathbb Z)$-transforms of $z$. Therefore $|w|\ge 1$. Together with the strip inequality, this gives $w\in\mathcal F$. Thus every point of $\mathbb H$ is $SL_2(\mathbb Z)$-equivalent to a point of $\mathcal F$.[/step]

[guided]At this stage the point $w$ is already in the vertical strip. The only remaining condition for $w\in\mathcal F$ is $|w|\ge 1$. We prove this using the same maximization principle. Assume that $|w|<1$. Applying \begin{align*} S=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \end{align*} gives \begin{align*} S w=-\frac{1}{w}. \end{align*} Writing $w=u+iv$ with $v=\operatorname{Im}(w)>0$, we compute \begin{align*} -\frac{1}{w} = -\frac{\overline w}{|w|^2} = \frac{-u+iv}{|w|^2}, \end{align*} so \begin{align*} \operatorname{Im}(Sw)=\frac{v}{|w|^2} =\frac{\operatorname{Im}(w)}{|w|^2}. \end{align*} If $|w|<1$, then $|w|^2<1$, and therefore \begin{align*} \operatorname{Im}(Sw)>\operatorname{Im}(w). \end{align*} But $S\in SL_2(\mathbb Z)$, so $Sw$ is another $SL_2(\mathbb Z)$-transform of the original point $z$. This contradicts the choice of $w$ as a transform with maximal imaginary part. Hence $|w|\ge 1$, and therefore \begin{align*} w\in \{z\in\mathbb H: |z|\ge 1,\ -\tfrac12\le \operatorname{Re}(z)\le \tfrac12\}=\mathcal F. \end{align*}[/guided]

[step:Prove that equivalent interior points are equal]Let $z,w\in\operatorname{int}(\mathcal F)$ and suppose \begin{align*} w=\gamma z,\qquad \gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z). \end{align*} Since $z,w\in\operatorname{int}(\mathcal F)$, we have \begin{align*} |z|>1,\qquad |\operatorname{Re}(z)|<\frac12,\qquad |w|>1,\qquad |\operatorname{Re}(w)|<\frac12. \end{align*} Apply the imaginary-part formula to $\gamma$ and to $\gamma^{-1}$: \begin{align*} \operatorname{Im}(w)=\frac{\operatorname{Im}(z)}{|cz+d|^2}, \qquad \operatorname{Im}(z)=\frac{\operatorname{Im}(w)}{|-cw+a|^2}. \end{align*} Thus \begin{align*} |cz+d|\,|-cw+a|=1. \end{align*} We use the following estimate. If $\zeta=X+iY\in\mathcal F$, where $X:=\operatorname{Re}(\zeta)$ and $Y:=\operatorname{Im}(\zeta)$, and if $m,n\in\mathbb Z$ with $m\neq 0$, then \begin{align*} |m\zeta+n|^2 &=m^2|\zeta|^2+2mnX+n^2 \\ &\ge m^2+n^2-|mn|. \end{align*} Here we used $|\zeta|\ge 1$ and $|X|\le \tfrac12$. For nonnegative integers $M:=|m|$ and $N:=|n|$, the integer $M^2+N^2-MN$ is at least $1$ because $M\ge 1$. Thus $|m\zeta+n|\ge 1$. If $\zeta\in\operatorname{int}(\mathcal F)$, then the inequalities $|\zeta|>1$ and $|X|<\tfrac12$ make the same estimate strict, so $|m\zeta+n|>1$. Apply this strict estimate first to $\zeta=z$ and $(m,n)=(c,d)$, and then to $\zeta=w$ and $(m,n)=(-c,a)$. If $c\neq 0$, then \begin{align*} |cz+d|>1, \qquad |-cw+a|>1, \end{align*} which contradicts $|cz+d|\,|-cw+a|=1$. Hence $c=0$. With $c=0$, the determinant condition gives $ad=1$, so $a=d=\pm1$. The transformation becomes \begin{align*} w=\frac{az+b}{d}=z+\frac{b}{d}. \end{align*} Since $b/d\in\mathbb Z$ and both real parts lie in the open interval $(-\tfrac12,\tfrac12)$, the only possible integer translation is $b/d=0$. Therefore $w=z$.[/step]

[guided]We now show that the interior of $\mathcal F$ contains exactly one representative from each orbit. Let \begin{align*} z,w\in\operatorname{int}(\mathcal F) \end{align*} and assume that \begin{align*} w=\gamma z,\qquad \gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z). \end{align*} Interior membership means that all defining inequalities are strict: \begin{align*} |z|>1,\qquad |\operatorname{Re}(z)|<\frac12, \end{align*} and the same inequalities hold for $w$. The imaginary-part formula gives \begin{align*} \operatorname{Im}(w)=\frac{\operatorname{Im}(z)}{|cz+d|^2}. \end{align*} Applying the same formula to \begin{align*} \gamma^{-1}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix} \end{align*} and the point $w$ gives \begin{align*} \operatorname{Im}(z)=\frac{\operatorname{Im}(w)}{|-cw+a|^2}. \end{align*} Multiplying these two identities yields \begin{align*} |cz+d|^2|-cw+a|^2=1. \end{align*} The geometry of the fundamental region is encoded in a direct estimate. Let $\zeta=X+iY\in\operatorname{int}(\mathcal F)$, where $X:=\operatorname{Re}(\zeta)$ and $Y:=\operatorname{Im}(\zeta)$, and let $m,n\in\mathbb Z$ with $m\neq 0$. Since $\zeta$ is in the interior of $\mathcal F$, we have \begin{align*} |\zeta|>1, \qquad |X|<\frac12. \end{align*} Compute \begin{align*} |m\zeta+n|^2 &=(mX+n)^2+m^2Y^2 \\ &=m^2(X^2+Y^2)+2mnX+n^2 \\ &=m^2|\zeta|^2+2mnX+n^2. \end{align*} The term $2mnX$ is bounded below by $-|mn|$ because $|X|<\tfrac12$, and the inequality is strict when $mn\neq 0$. Also $m^2|\zeta|^2>m^2$ because $m\neq 0$ and $|\zeta|>1$. Therefore \begin{align*} |m\zeta+n|^2>m^2+n^2-|mn|. \end{align*} Putting $M:=|m|\in\mathbb N$ and $N:=|n|\in\mathbb N\cup\{0\}$, the integer \begin{align*} M^2+N^2-MN \end{align*} is at least $1$: if $N=0$, it equals $M^2\ge 1$; if $N\ge 1$, then $M^2+N^2-MN=(M-N)^2+MN\ge 1$. Hence \begin{align*} |m\zeta+n|>1. \end{align*} This is the precise replacement for the informal picture of non-overlapping translated disks. Apply the estimate with $\zeta=z$ and $(m,n)=(c,d)$. If $c\neq0$, then $|cz+d|>1$. Apply it again with $\zeta=w$ and $(m,n)=(-c,a)$. If $c\neq0$, then $|-cw+a|>1$. Their product would be greater than $1$, contradicting \begin{align*} |cz+d|\,|-cw+a|=1. \end{align*} Hence $c=0$. When $c=0$, the determinant condition $ad-bc=1$ reduces to $ad=1$, so $a=d=1$ or $a=d=-1$. In both cases \begin{align*} w=\frac{az+b}{d}=z+\frac{b}{d}, \end{align*} where $b/d\in\mathbb Z$. Since $\operatorname{Re}(z)$ and $\operatorname{Re}(w)$ both lie strictly between $-\tfrac12$ and $\tfrac12$, no nonzero integer translation can carry one to the other. Thus $b/d=0$, and consequently $w=z$.[/guided]

[step:Identify the boundary pairings]The boundary of $\mathcal F$ consists of the vertical sides $\operatorname{Re}(z)=\pm\tfrac12$ with $|z|\ge 1$ and the circular arc $|z|=1$ with $-\tfrac12\le \operatorname{Re}(z)\le\tfrac12$. On the vertical sides, the map \begin{align*} T:\mathbb H&\to\mathbb H\\ z&\mapsto z+1 \end{align*} sends the left side $\operatorname{Re}(z)=-\tfrac12$ to the right side $\operatorname{Re}(z)=\tfrac12$ and preserves imaginary part. Thus the two vertical sides are paired by $T$ and $T^{-1}$. On the circular arc $|z|=1$ with $-\tfrac12\le \operatorname{Re}(z)\le \tfrac12$, the map \begin{align*} S:\mathbb H&\to\mathbb H\\ z&\mapsto -\frac1z \end{align*} satisfies, because $\overline z=1/z$ on $|z|=1$, \begin{align*} S z=-\overline z. \end{align*} Hence $S$ reflects the arc across the imaginary axis, pairing the two halves of the circular boundary and fixing $i$. It remains to exclude any further boundary identifications. Let $z,w\in\mathcal F$ and suppose $w=\gamma z$ for \begin{align*} \gamma:=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z). \end{align*} As in the interior argument, \begin{align*} |cz+d|\,|-cw+a|=1. \end{align*} If $c=0$, then $ad=1$, so $a=d=1$ or $a=d=-1$, and the transformation is \begin{align*} w=z+\frac{b}{d}, \end{align*} where $b/d\in\mathbb Z$. Since both real parts lie in the closed interval $[-\tfrac12,\tfrac12]$, the only possibilities are $b/d=0$, which gives $w=z$, or $b/d=1$ with $\operatorname{Re}(z)=-\tfrac12$ and $\operatorname{Re}(w)=\tfrac12$, or $b/d=-1$ with the two sides reversed. These are exactly the vertical side identifications generated by $T$. Assume now that $c\neq0$. We use the equality cases in the estimate from the interior argument. Let $\zeta=X+iY\in\mathcal F$, where $X:=\operatorname{Re}(\zeta)$ and $Y:=\operatorname{Im}(\zeta)$, and let $m,n\in\mathbb Z$ with $m\neq0$. Put $M:=|m|\in\mathbb N$ and $N:=|n|\in\mathbb N\cup\{0\}$. Then \begin{align*} |m\zeta+n|^2 &=m^2|\zeta|^2+2mnX+n^2 \\ &\ge M^2+N^2-MN \\ &\ge 1. \end{align*} Equality holds only if all inequalities used here are equalities. Thus $|\zeta|=1$, and either $N=0$ with $M=1$, or $M=N=1$ with $X=-\operatorname{sgn}(mn)/2$. Consequently, if $|m\zeta+n|=1$, then either $(m,n)=(\pm1,0)$ and $\zeta$ lies on the circular arc, or $|m|=|n|=1$ and $\zeta$ is one of the two endpoints \begin{align*} \rho_+ := \frac12+\frac{\sqrt3}{2}i, \qquad \rho_- := -\frac12+\frac{\sqrt3}{2}i. \end{align*} Since $|cz+d|\ge1$ and $|-cw+a|\ge1$ while their product is $1$, both factors are equal to $1$. Applying the equality classification to $(\zeta,m,n)=(z,c,d)$ shows that either $d=0$ and $|c|=1$, or $z$ is an endpoint of the circular arc. Applying it to $(\zeta,m,n)=(w,-c,a)$ shows the same alternative for $a$ and $w$. If $d=0$, then the determinant equation $ad-bc=1$ becomes $-bc=1$, so $|b|=|c|=1$. The associated fractional linear transformation is \begin{align*} \gamma z=\frac{az+b}{cz}=\frac{a}{c}+\frac{b}{cz}. \end{align*} Because $d=0$ and $|c|=1$, the equality case for $z$ gives $|z|=1$. Also $-bc=1$, so $b/c=-1$, and hence \begin{align*} \gamma z=\frac{a}{c}-\frac1z. \end{align*} For $\gamma z$ to lie in $\mathcal F$, the real part can remain in $[-\tfrac12,\tfrac12]$ only after the possible integer translation $a/c\in\mathbb Z$; equivalently, this identification is obtained from the circular pairing $S:z\mapsto -1/z$ followed, if necessary, by a vertical side translation generated by $T$. Thus it is generated by $S$ and $T$. If $d\neq0$, the equality classification for $|cz+d|=1$ forces $|c|=|d|=1$ and forces $z$ to be an endpoint, with $\operatorname{Re}(z)=-\operatorname{sgn}(cd)/2$. Similarly, $|a|=|c|=1$ and $w$ is an endpoint. The endpoints are already identified by the adjacent boundary pairings: $S\rho_+=\rho_-$, $T\rho_- = \rho_+$, and the reversed identifications are generated by $S^{-1}=S$ and $T^{-1}$. Hence these endpoint cases introduce no generator beyond $S$ and $T$. Therefore every identification among distinct boundary points of $\mathcal F$ is generated by the vertical side pairing $T$ and the circular arc pairing $S$. This completes the proof.[/step]

[guided]We must prove two things about the boundary: first, that the announced maps really identify boundary points, and second, that no other identifications occur. The vertical side pairing is direct. Define \begin{align*} T:\mathbb H&\to\mathbb H\\ z&\mapsto z+1. \end{align*} If $\operatorname{Re}(z)=-\tfrac12$, then $\operatorname{Re}(Tz)=\tfrac12$, and $\operatorname{Im}(Tz)=\operatorname{Im}(z)$. Hence $T$ pairs the left vertical side of $\mathcal F$ with the right vertical side, and $T^{-1}$ gives the reverse pairing. The circular pairing is also explicit. Define \begin{align*} S:\mathbb H&\to\mathbb H\\ z&\mapsto -\frac1z. \end{align*} If $|z|=1$, then $1/z=\overline z$, so \begin{align*} S z=-\overline z. \end{align*} Thus $S$ reflects the circular arc across the imaginary axis. In particular it pairs the two halves of the arc and fixes the midpoint $i$. Now suppose an arbitrary modular transformation identifies two points of $\mathcal F$. Let \begin{align*} z,w\in\mathcal F, \qquad w=\gamma z, \qquad \gamma:=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z). \end{align*} Applying the imaginary-part formula to $\gamma$ and to \begin{align*} \gamma^{-1}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix} \end{align*} gives \begin{align*} \operatorname{Im}(w)=\frac{\operatorname{Im}(z)}{|cz+d|^2}, \qquad \operatorname{Im}(z)=\frac{\operatorname{Im}(w)}{|-cw+a|^2}. \end{align*} Multiplying yields \begin{align*} |cz+d|\,|-cw+a|=1. \end{align*} This identity forces equality in the same estimate that proved interior uniqueness. First consider $c=0$. Then $ad=1$, so $a=d=1$ or $a=d=-1$, and \begin{align*} w=z+\frac{b}{d} \end{align*} with $b/d\in\mathbb Z$. Since $\operatorname{Re}(z)$ and $\operatorname{Re}(w)$ both lie in $[-\tfrac12,\tfrac12]$, an integer translation can preserve membership in the strip only if it is $0$, $1$, or $-1$. Translation by $0$ gives the same point. Translation by $1$ sends the left side to the right side, and translation by $-1$ reverses that pairing. These are exactly the identifications generated by $T$. It remains to handle $c\neq0$. We record the exact equality cases. Let $\zeta=X+iY\in\mathcal F$, with $X:=\operatorname{Re}(\zeta)$ and $Y:=\operatorname{Im}(\zeta)$, and let $m,n\in\mathbb Z$ with $m\neq0$. Put $M:=|m|\in\mathbb N$ and $N:=|n|\in\mathbb N\cup\{0\}$. Since $|\zeta|\ge1$ and $|X|\le\tfrac12$, \begin{align*} |m\zeta+n|^2 &=m^2|\zeta|^2+2mnX+n^2 \\ &\ge M^2+N^2-MN \\ &\ge 1. \end{align*} Equality is restrictive. It requires $|\zeta|=1$. It also requires equality in $2mnX\ge -|mn|$, so if $mn\neq0$ then $X=-\operatorname{sgn}(mn)/2$. Finally it requires $M^2+N^2-MN=1$. This last integer equation gives either $M=1,N=0$, or $M=N=1$. Therefore $|m\zeta+n|=1$ occurs only in two ways: either $(m,n)=(\pm1,0)$ and $\zeta$ lies on the circular arc, or $|m|=|n|=1$ and $\zeta$ is one of the two endpoints \begin{align*} \rho_+ := \frac12+\frac{\sqrt3}{2}i, \qquad \rho_- := -\frac12+\frac{\sqrt3}{2}i. \end{align*} Apply this to the two factors $|cz+d|$ and $|-cw+a|$. Since both factors are at least $1$ and their product is $1$, both must equal $1$. If $d=0$, then the determinant equation gives $-bc=1$, so $|b|=|c|=1$. The transformation has the form \begin{align*} \gamma z=\frac{az+b}{cz}=\frac{a}{c}+\frac{b}{cz}. \end{align*} Because $b/c=-1$, this becomes \begin{align*} \gamma z=\frac{a}{c}-\frac1z. \end{align*} The term $z\mapsto -1/z$ is exactly the circular pairing $S$, and $a/c\in\mathbb Z$ is an integer translation, hence a power of $T$. Thus this case is generated by $S$ and $T$. If $d\neq0$, the equality case for $|cz+d|=1$ forces $|c|=|d|=1$ and forces $z$ to be an endpoint of the circular arc. The equality case for $|-cw+a|=1$ similarly forces $w$ to be an endpoint. But the endpoints are already related by the adjacent side and arc pairings: \begin{align*} S\rho_+=\rho_-, \qquad T\rho_- = \rho_+. \end{align*} The reverse identifications are generated by $S^{-1}=S$ and $T^{-1}$. Therefore endpoint identifications introduce no generator beyond $S$ and $T$. Combining the $c=0$ and $c\neq0$ cases, every boundary identification among distinct points of $\mathcal F$ is generated by the vertical side pairing $T$ and the circular arc pairing $S$.[/guided]