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 ?

enter image description here

  • $\begingroup$ To understand this proof you'll need at least theorem 1.2 from Farb Margalit, the classification of isometries of $\mathbb{H}^2$ (the isometry $\varphi$ the proof talks about is a hyperbolic isometry with axis the geodesic joining the endpoints of $\widetilde{\alpha}$ and $\widetilde{\beta}$), basic covering space theory and the obvious fact that if two curves are the boundary components of some annulus then they are isotopic. $\endgroup$
    – Saúl RM
    Commented Jan 18, 2022 at 11:27
  • $\begingroup$ @SaúlRodríguezMartín I read this theorem. But I can't understand why There is a hyperbolic isometry $\phi$ that leaves $\widetilde{\alpha}$ and $\widetilde{\beta}$ invariant and acts by translation on these lifts ? Why? $\endgroup$
    – T566y65tt
    Commented Jan 18, 2022 at 11:33
  • $\begingroup$ @SaúlRodríguezMartín .why The quotient $R^{\prime}=$ $R /\langle\phi\rangle$ is an annulus ? $\endgroup$
    – T566y65tt
    Commented Jan 18, 2022 at 11:36
  • $\begingroup$ @SaúlRodríguezMartín . What is $R /\langle\phi\rangle$ ? What is $ \langle\phi\rangle$ ? $\endgroup$
    – T566y65tt
    Commented Jan 18, 2022 at 11:44
  • $\begingroup$ You could try my lecture notes here: dpmms.cam.ac.uk/~hjrw2/RS%20lectures.pdf . I wrote an updated version in the last few months -- I have a feeling one of the updates concerns this proof -- which I'll try to post in the next few days. $\endgroup$
    – HJRW
    Commented Jan 18, 2022 at 15:10

1 Answer 1


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 $D / (c_0 \sim c_1)$

  • $\begingroup$ So my shape is true ? $\endgroup$
    – T566y65tt
    Commented Jan 18, 2022 at 18:07
  • $\begingroup$ Neither of the pictures at the end of your post are an annulus. So, no? $\endgroup$
    – Sam Nead
    Commented Jan 18, 2022 at 18:40
  • $\begingroup$ Why the case $\chi(S)=0$ is similar? In this case we have euclidean surface but I don't know why we have similar way ? $\endgroup$
    – T566y65tt
    Commented Jan 19, 2022 at 8:02
  • $\begingroup$ Much of the argument is similar. However we will need to replace the hyperbolic geometry arguments with Euclidean ones. $\endgroup$
    – Sam Nead
    Commented Jan 19, 2022 at 22:11
  • $\begingroup$ is there any reference about your proof ? $\endgroup$
    – T566y65tt
    Commented Jan 22, 2022 at 22:25

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