In the " A Primer on Mapping Class Groups Benson Farb and Dan 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.

how we can prove the case $\chi(S)=0$ and the case $\chi(S)>0$ ? why The annulus $R^{\prime \prime}$ between $\alpha$ and $\beta$ gives the desired isotopy ? How we can prove $R^{\prime \prime}$ desired isotopy ?

I think if $\chi(S)=0$ then $2-2g-(b+n)=0$ so we have two case $g=0,1$ then we have a surface with $b+n=2,0$ then $\alpha$ and $\beta$ are isotopic.


1 Answer 1


Once you have found an annulus $R \subset S$ whose two boundary components are $\alpha$ and $\beta$, by definition of "annulus" there exists a homeomorphism $H : S^1 \times [0,1] \to R$. The composition $$S^1 \times [0,1] \xrightarrow{H} R \hookrightarrow S $$ then defines an isotopy in $S$ from $\alpha$ to $\beta$.

If $\chi(S)=0$ the surface $S$ is a torus or annulus, and we can then obtain the desired annulus $R$ using a Euclidean structure on $S$ in a manner similar to how the hyperbolic structure is used when $\chi(S)<0$.

In the case $\chi(S)>0$ the surface $S$ is a sphere or disc, in which case by the Schönflies theorem no essential simple closed curves exist and so the proposition is vacuously true.

  • $\begingroup$ If $\chi(S)=0$ why $S$ is torus? Why we have not case $g=0$ ? or sphere? $\endgroup$
    – T566y65tt
    Jan 16 at 17:22
  • $\begingroup$ The Euler characteristic of a sphere is equal to $2$. More generally, the Euler chacteristic of a closed, oriented surface of genus $g$ is equal to $2-2g$. $\endgroup$
    – Lee Mosher
    Jan 16 at 17:45
  • $\begingroup$ $b$ is the number of boundary components. One way to obtain a noncompact surface from a compact surface $S$ is to remove $n$ points from the interior of $S$ We can therefore specify our surfaces by the triple $(g, b, n)$. We will denote by $S_{g, n}$ a surface of genus $g$ with $n$ punctures and empty boundary; such a surface is homeomorphic to the interior of a compact surface with $n$ boundary components. Recall that the Euler characteristic of a surface $S$ is $$ \chi(S)=2-2 g-(b+n) . $$ $\endgroup$
    – T566y65tt
    Jan 16 at 18:32
  • $\begingroup$ So in this book surface is compact so we have $n=0$ but why $b=0$ ? $\endgroup$
    – T566y65tt
    Jan 16 at 18:33
  • 1
    $\begingroup$ Schönflies Theorem $\endgroup$
    – Lee Mosher
    Jan 16 at 23:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.