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Take two solid tori $V_1,V_2$, each with a meridian-longitude basis $({\mu }_i,{\lambda }_i)$ on its boundary. Use opposite boundary-basis orientations on the two sides, so that a determinant-one coefficient matrix describes an orientation-reversing gluing map. For $A\in SL(2,\mathbb{Z})$ choose a torus homeomorphism $f:\partial V_2\to \partial V_1$ inducingand define $M_A=V_1{\cup }_f{V}_2$. The linear map on ${\mathbb{R}}^2/{\mathbb{Z}}^2$ realizes the matrix. Torus homeomorphisms with the same action on first homology are isotopic, and an isotopy extends over a boundary collar, so the resulting three-manifold depends only on $A$. Specifying opposite boundary conventions is necessary: gluing two identically oriented boundary bases by a determinant-one map would not describe the usual oriented gluing.

We now identify this gluing directly in the quotient from part (a). Split $S^3$ into the two invariant solid tori $W_1=\{|z_2|\le 1/\sqrt{2}\}$ and $W_2=\{|z_1|\le 1/\sqrt{2}\}$. On $W_1$ write $z_1=\sqrt{1-|w{|}^2}{e}^{2\pi ix}$, $z_2=w$. The quotient coordinatesidentify $W_1/⟨g⟩$ with $S^1\times D^2$. They are invariant under $g$; choosing a $p$th root of $u$ reconstructs $(x,w)$, and different choices are exactly the same cyclic orbit. Thus this is a genuine solid-torus coordinate model.

On the common boundary, measure arguments in turns as $(x,y)$. Its quotient lattice isThe meridian and longitude of the first quotient torus can be chosen as ${\mu }_1=(0,1)$ and ${\lambda }_1=(1/p,q/p)$. They form a lattice basis, since $(1,0)=p{\lambda }_1-q{\mu }_1$. The meridian of the second torus is precisely that $(1,0)$ loop. Consequently

The determinant condition is $qr+ps=-1$. Put $k=-r$, so $kq\equiv 1(\mathrm{mod}p)$. On $W_2$, the analogous quotient coordinates use $e^{2\pi ipy}$ and $e^{-2\pi iky}{z}_1$, and a longitude is $(k/p,1/p)$. Choose its negative for the second boundary-basis convention. Thensince $ps+qr=-1$. Thus the two quotient tori are glued by exactly $A$. This proves the genus-one gluing model of a lens space andDifferent choices of $r,s$ satisfying the determinant condition differ by $(s,r)\mapsto (s,r)+j(-q,p)$. This changes $f$ by a meridional Dehn twist on $V_2$, which extends over $V_2$: in solid-torus coordinates it is $(z,e^{i\theta })\mapsto (e^{ij\theta }z,e^{i\theta })$. Therefore there is no hidden dependence on that choice of longitude.