Isotopy classes of embedded closed curves in a torus Can anyone explain to me why the Isotopy classes of embedded closed curves in a torus are classified by “slopes” $r\in\mathbb{Q}\cup \{\infty\}$? Why can we identify (up to isotopy) simple closed curves ($(p,q)$-curves) in a Torus to $\mathbb{Q}\cup \{\infty\}$? I thought this is because of the fact that the fundamental group of the Torus is $\mathbb{Z}\times\mathbb{Z}$ but I am not really sure. I have seen the above fact being used several times. (see Ulrich  page 2, Hatcher page 1).
 A: For simplicity, I'm going to assume we're in the smooth category or the piecewise linear (PL) category.  I'll leave the topological category for you, someone else, or later.
Let $C$ be a simple closed curve in a torus $T$.  Fix a basepoint along $C$, and say $(0,0)$ is the corresponding basepoint in the universal cover $\mathbb{R}^2$ of $C$.  First of all, isotopies of oriented simple closed curves are homotopies, so the corresponding element of $\pi_1(T)$ is an invariant of the isotopy class of the oriented simple closed curve.  For unoriented simple closed curves, the invariant instead lies in $\pm\pi_1(T)$, written this way since the fundamental group is abelian.
If $C$ is separating, meaning $T-C$ is disconnected, then $C$ gives $T$ as a connect sum of two surfaces, one of which is a sphere (by the classification of surfaces).  That means one of the connected components of $T-C$ is a disk, so $C$ is nullhomotopic, corresponding to $(0,0)$ in $\pm\pi_1(T)$.  Let's say in your question we ignore nullhomotopic simple closed curves.  (Alternatively, separating means that $C$ is nullhomologous because each component of $T-C$ represents a surface whose boundary is $C$.  Pair that with $H_1(T)\cong\pi_1(T)$, since the fundamental group is already abelian.)
If $C$ is non-separating, then since $T$ is orientable there is a path from one side of $C$ to the other.  In particular, take a closed regular neighborhood $\overline{\nu(C)}\subset T$ of $C$, which is a cylinder (annulus), and take a path $P$ in $T-\nu(C)$ between the two boundary components of $\overline{\nu(C)}$.  The boundary of $\overline{\nu(C)}\cup\overline{\nu(P)}$ is a separating circle, again demonstrating $T$ as a connect sum.  Since $C$ is homotopically non-trivial, the $C$-side of the boundary is a torus connect summand, and the other side is a disk (again, by the classification of surfaces).  One gets an isomorphism between $\pi_1(T)$ and $\mathbb{Z}^2$ from this where one of the orientations of $C$ corresponds to $(1,0)$.
A consequence of this is that $p$ and $q$ are coprime.  Every automorphism of $\mathbb{Z}^2$ can be written as a $2\times 2$ invertible matrix of integer entries with determinant $\pm 1$.  The determinant of the matrix is divisible by the gcd of $p$ and $q$, and so $p$ and $q$ are coprime.
Let $C_{p,q}$ be the image in $T$ of the straight-line path from $(0,0)$ to $(p,q)$ in the universal cover.  If $p,q$ are coprime, then the image in $T$ is a simple closed curve.  Therefore, for unoriented simple closed curves, the invariant is in $\{\pm (p,q)\in\pm\pi_1(T):\gcd(p,q)=1\}$, and every such invariant value is obtainable from some simple closed curve.  By using the map $\pm(p,q)\mapsto p/q$, with $1/0=\infty$, we can write this invariant of isotopy classes of unoriented non-separating simple closed curves, the slope, as an element of $\mathbb{Q}\cup\{\infty\}$.
Now, why is this a complete invariant?  Take a non-separating simple closed curves $C$ in $T$.  We may as well take a self-homeomorphism of $T$ so that without loss of generality we may assume $C$ has a slope of $1/0$.  We will show that $C$ is isotopic to $C_{1,0}$.  By a standard appeal to transversality, we may assume $C$ intersects $C_{1,0}$ transversely at finitely many points, since it is isotopic to one that does.
We're going to show that $C$ is isotopic to a curve that does not intersect $C_{1,0}$.  In $T-C_{1,0}$, $C-C_{1,0}$ is a collection of loops and arcs.  If there are any loops, then there are no arcs and hence no intersection, so consider the case of no loops.  Arcs come in two varieties: those that go from one side of $C_{1,0}$ to the other ("through-arcs") and those that go from one side of $C_{1,0}$ to the same side ("back-arcs").  Consider the regions of $T-(C\cup C_{1,0})$.  If there is a back-arc, then there is a disk region bounded only by that back-arc and an arc from $C_{1,0}$, by taking an "inner-most" back-arc according to some partial order.  Use this disk to isotope $C$ to remove two points of intersection.  By induction, there are no back-arcs and only through-arcs.  If there were only through-arcs, though, then the slope of $C$ couldn't be $1/0$, but instead $a/b$ where $b$ is the number of through-arcs.
Thus, $C$ is isotopic to a loop in the annulus $T-C_{1,0}$, and by the theory of isotopy classes of separating simple closed curves in an annulus, $C$ is isotopic to the core circle, and therefore to $C_{1,0}$.
A: Let $p$ be the number of times the curve loops around the meridian of the torus, and $q$ be the number of times the curve loops around the equator of the torus. Then $p/q \in \Bbb{Q} \cup \{ \infty \}$ describes the curve up to isotopy. We include $\infty$ since we may have $q=0$.
A: I believe it's because curves with rational slope can be seen on the unit square with opposite sides identified (a typical way of realizing the torus) to be precisely the simple closed curves (that is they circle around and land on top of themselves);  contrasted with those with irrational slope, which are not closed and are in fact dense in the torus (this is an exercise in Guilleman and Pollack).
A: To add to the existing answers, the correspondence with rational numbers is not made for just any element of $\pi_1(\Bbb T^2)$, it is made for the subset of those elements that can be represented by embedded curves. 
In that case, the algebraic number of times a curve hits the longitude and the meridian, say $p$ and $q$, have to be coprime. This is why the map $(m,n)\mapsto\frac mn$ restricted to this class of curves is injective. 
The reason why they have to be coprime is explained in this answer by Lee Mosher. Note that if $p$ and $q$ are not coprime then the homology class of the curve is a multiple of another homology class. Standard generators of $H_1(\Bbb T^2)$ are not multiples of other homology classes.
A: $\bullet$ Suppose $(p,q)$ is a simple torus curve. There exists a closed curve $(a,b)$ that intersects $(p,q)$ exactly once. Therefore, $|ap-bq|=1$ (see here), which shows that $pb-qa=\pm 1$, so $\gcd(p,q)=1$.
$\bullet$ Suppose that $\gcd(p,q)=1$. We know that the fundamental group of the Torus is generated by $[\nu]$ and $[\mu]$, where $\mu$ is the $(1,0)$-curve and $\nu $ is the $(0,1)$-curve (we consider them as loops based at the origin). The element $p[\mu]+q[\nu]$ is an element of the fundamental group and it is primitive (due to the fact that $\gcd(p,q)=1$). $[(p,q)]\in\pi_1(T)$ is not a multiple of any elements. Therefore, the $(p,q)$-curve cannot intersect itself. It follows that the $(p,q)$-curve must be a simple curve.
Now, it is clear that there is a bijection between the set of simple closed curves on the Torus and $\mathbb{Q}\cup\{\infty\}$.
