Primitive pairs define unique unoriented slopes on a torus

Let $p,q\in \mathbb{Z}$ satisfy $\gcd (p,q)=1$. Bézout's identity gives $r,s\in \mathbb{Z}$ with $ps-qr=1$. The matrix with columns $(p,q)$ and $(r,s)$ lies in $\mathrm{S}\mathrm{L}(2,\mathbb{Z})$, so it induces a homeomorphism of ${\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2$. The image of the first coordinate circle is embedded, essential, and represents $p[\mu ]+q[\lambda ]$. Reversing its parameter negates both coefficients, so the unoriented curve is unchanged.

If $\gamma$ is an essential simple closed curve on $T$, it is nonseparating; cutting along it gives an annulus. An arc joining the two annulus boundary circles closes to a simple curve $\delta$ meeting $\gamma$ once, so the algebraic intersection is $[\gamma ]\cdot [\delta ]=\pm 1$. Writing $[\gamma ]=p[\mu ]+q[\lambda ]$ and $[\delta ]=a[\mu ]+b[\lambda ]$ gives $pb-qa=\pm 1$ because the ordered basis has intersection form $\pm 1$. Any common divisor of $p,q$ would divide $pb-qa$, so $\gcd (p,q)=1$.

Suppose ${\gamma }_0,{\gamma }_1$ are essential simple closed curves whose primitive classes agree up to sign. Reverse one orientation if necessary so their classes agree, and isotope them to be transverse. Let $v\in {\mathbb{Z}}^2$ be their common primitive class in the coordinates of [A1]. If they intersect, choose lifts ${\tilde{\gamma }}_0,{\tilde{\gamma }}_1\subset {\mathbb{R}}^2$ through one lift $x$ of an intersection point. Each chosen lift is preserved by the same nonzero deck translation $T_v$, because its projected curve represents $v$. Hence both $x$ and $T_v(x)$ lie on this pair of lifts and are distinct. By [F1] the curves therefore form an embedded bigon in $T$ whose interior misses them. Isotoping one curve across this bigon removes exactly two intersections and creates none. The transverse intersection set is finite, so repetition terminates and makes the curves disjoint. Cutting $T$ along ${\gamma }_0$ gives an annulus, and the disjoint essential curve ${\gamma }_1$ is parallel to its core, so the two curves cobound a subannulus. Isotopy across that annulus carries ${\gamma }_0$ to ${\gamma }_1$. Thus the primitive class modulo simultaneous sign determines exactly one unoriented isotopy class.