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In the book by Benson Farb and Dan Margalit A primer on mapping class groups, Princeton Mathematical Series 49. Princeton, NJ: Princeton University Press (ISBN 978-0-691-14794-9/hbk; 978-1-400-83904-9/ebook), pp. xiv+492 (2011), MR2850125, Zbl 1245.57002,
we have the following statement:

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.

Questions

  • Why is there a hyperbolic isometry $\phi$ leaving $\widetilde{\alpha}$ and $\widetilde{\beta}$ invariant and acting by translation on these lifts or $\phi (\widetilde{\beta})=\widetilde{\beta}$ and $\phi (\widetilde{\alpha})=\widetilde{\alpha}$?
  • Why is the quotient $R^{\prime}=$ $R /\langle\phi\rangle$ an annulus? How we can show this is an annulus? I can't understand what is "$\langle\phi\rangle$".
  • Why is the covering map $R^{\prime} \rightarrow R^{\prime \prime}$ single-sheeted on the boundary?
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1 Answer 1

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There is a great deal of overlap between this question and the following questions asked by the same person:

However, the exact questions asked are a bit different, so the answers are (a little bit) different as well.


We will work under the assumptions that $S$ is compact, connected, without boundary, oriented and of negative Euler characteristic. All of these assumptions can be loosened, of course. But when dealing with the theory for the first time, this case is the one to focus on.

Question 1: Due to our assumptions we can equip $S$ with a complete hyperbolic metric of finite area. Thus $\pi_1(S)$ acts on the hyperbolic plane. Again due to our assumptions, all non-trivial elements of $\pi_1(S)$ act hyperbolically. Since $\alpha$ and $\beta$ are essential, they give (the same) non-trivial element $\phi$ of $\pi_1(S)$. It is this element of the fundamental group that the proof is referring to.

Question 2: Since $S$ is oriented, all elements of $\pi_1(S)$ act on the hyperbolic plane preserving orientation. (If $S$ was not orientable, then glide reflections are possible.) Let $A_\phi$ be the geodesic axis of $\phi$. Let $P$ be any geodesic perpendicular to $A_\phi$. Then $L$ and $\phi(L)$ cobound a "strip" $P$ in the hyperbolic plane. Gluing $L$ to $\phi(L)$ via $\phi$ gives a quotient $P / \phi$ which is an annulus. This annulus contains an embedded copy of $R'$. So $R'$ is an annulus.

Also, $\langle \phi \rangle$ is the cyclic subgroup (of $\pi_1(S)$, essentially) generated by $\phi$.

Question 3: By path lifting and the definition of $\phi$.

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