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.