In the Primer mapping class group by farb Margalit. We have :
Proposition 1.10 Let $\alpha$ and $\beta$ be two essential simple closed curves in a surface $S$. Then $\alpha$ is isotopic to $\beta$ if and only if $\alpha$ is homotopic to $\beta$.
Proof. One direction is vacuous since an isotopy is a homotopy. So suppose that $\alpha$ is homotopic to $\beta$. We immediately have that $i(\alpha, \beta)=0$. By performing an isotopy of $\alpha$, we may assume that $\alpha$ is transverse to $\beta$. If $\alpha$ and $\beta$ are not disjoint, then by the bigon criterion they form a bigon. A bigon prescribes an isotopy that reduces intersection. Thus we may remove bigons one by one by isotopy until $\alpha$ and $\beta$ are disjoint.
In the remainder of the proof, we assume $\chi(S)<0$; the case $\chi(S)=0$ is similar, and the case $\chi(S)>0$ is easy. Choose lifts $\widetilde{\alpha}$ and $\widetilde{\beta}$ of $\alpha$ and $\beta$ that have the same endpoints in $\partial \mathbb{H}^{2}$. There is a hyperbolic isometry $\phi$ that leaves $\widetilde{\alpha}$ and $\widetilde{\beta}$ invariant and acts by translation on these lifts. As $\widetilde{\alpha}$ and $\widetilde{\beta}$ are disjoint, we may consider the region $R$ between them. The quotient $R^{\prime}=$ $R /\langle\phi\rangle$ is an annulus; indeed, it is a surface with two boundary components with an infinite cyclic fundamental group. A priori, the image $R^{\prime \prime}$ of $R$ in $S$ is a further quotient of $R^{\prime}$. However, since the covering map $R^{\prime} \rightarrow R^{\prime \prime}$ is single-sheeted on the boundary, it follows that $R^{\prime} \approx R^{\prime \prime}$. The annulus $R^{\prime \prime}$ between $\alpha$ and $\beta$ gives the desired isotopy .
I can't understand this proof for $\chi(S)<0$ . Is there any reference for more details for this proposition? Or simple proof ? I read paper by "bear" in the original proof but this is more difficult.
I think we consider $ \langle \phi\rangle$ as discrate topological group acts on $R$ but why quotient $R^{\prime}=$ $R /\langle\phi\rangle$ is an annulus ? also I think
$ \langle \phi\rangle$ is isomorphic to infinite cyclic group $\mathbb Z$
Is this following shape true ?
Best Answer
Suppose that $S$ is a closed connected surface with negative Euler characteristic. Suppose that $H$ is the universal cover of $S$. Equipping $S$ with a hyperbolic metric we find that $H$ is isometric to the hyperbolic plane.
Exercise: Every non-trivial deck transformation of $H$ over $S$ is hyperbolic: that is, has exactly two fixed points at infinity.
Suppose that $\alpha$ is a simple closed curve. Suppose that $\alpha$ is essential: in particular it does not bound a disk in $S$. We choose a point $x_0$ of $\alpha$ to be the basepoint of $\pi_1(S, x_0)$. Let $y_0 \in H$ be any preimage of $x_0$.
Let $\phi : ([0, 1], \{0, 1\}) \to (S, x_0)$ be a parameterisation of $\alpha$. Note that $[\phi]$ is an element of $\pi_1(S, x_0)$. We lift $\phi$ via path lifting to obtain a simple arc $\phi_0 : ([0, 1], \{0, 1\}) \to (H, {y_0, y_1})$. Note that path lifting defines the point $y_1$. There is a unique deck transformation that takes $y_0$ to $y_1$. In fact, this is the image of $[\phi]$ in the deck group. We will abuse notation and simply call this deck transformation $[\phi]$.
Exercise: $y_1$ is not equal to $y_0$. Thus the deck transformation $[\phi]$ is a hyperbolic transformation of $H$.
Let $\langle \phi \rangle$ be the subgroup of the deck group generated by $[\phi]$. Define $\phi_k = [\phi]^k \cdot \phi_0$. Let $A = \cup \phi_k$ be the union of the translates of $\phi_0$. We call $A$ an elevation of $\alpha$ (as it is a union of lifts giving a copy of the universal cover of $\alpha$).
Exercise: $A$ is a quasi-geodesic in $H$.
Thus $A$ has two endpoints in $\partial H$, the ideal boundary of $H$.
Exercise: Suppose that $\beta$ is a loop in $S$ that is (freely) homotopic to $\alpha$. Then all components of the preimage of $\beta$, in $H$, are quasi-geodesics, and exactly one (call it $B$) shares the ideal endpoints of $A$.
We now require an additional hypothesis. We must suppose that $S$ is orientable. If we do not, then it is possible that $\alpha$ is a core curve of a Möbius band: in this case $\alpha$ meets all curves in its free homotopy class. (You may want to think about how this interacts with the bigon criterion.)
Suppose that $\beta$ is a simple closed curve which is freely homotopic to, and disjoint from, $\alpha$. Recall that $A$ and $B$ are the chosen quasi-geodesic elevations of $\alpha$ and $\beta$. We deduce that $A$ and $B$ are disjoint in $H$ and share their ideal points in $\partial H$. So they cobound an (infinite) strip $R$ in $H$. Choose an arc $c_0$ that embeds in $R$ and connects $x_0 \in A$ to a point of $B$.
Exercise: We can choose $c_0$ so that it is disjoint from $c_k = [\phi]^k \cdot c_0$.
Finally, we deduce that $c_0$ and $c_1 = [\phi] \cdot c_0$ cobound a disk $D$ contained in $R$. Thus $R / \langle \phi \rangle$ is homeomorphic to the annulus $D / (c_0 \sim c_1)$