# About isotopy and homotopy

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

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.

• If $\chi(S)=0$ why $S$ is torus? Why we have not case $g=0$ ? or sphere? Jan 16 at 17:22
• 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 $2-2g$. Jan 16 at 17:45
• $b$ 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 $$\chi(S)=2-2 g-(b+n) .$$ Jan 16 at 18:32
• So in this book surface is compact so we have $n=0$ but why $b=0$ ? Jan 16 at 18:33
• Schönflies Theorem Jan 16 at 23:23