# How to rigorously prove that simple closed curves on a surface are primitive closed curves ?

Let me first state the definitions :

A not-nullhomotopic closed curve / loop $c$ on an orientable surface $X,c:[0,1]\to X$ is called simple closed curve is $c|[0,1)$ is injective and [ $c(0)=c(1) ] ;$ A closed curve / loop $c$ is called primitive if in the fundamental group $\pi_1(X,c(1)),$ the homotopy class $[c]$ can NOT be written as $[c]= [\gamma]^n$ for some closed curve $\gamma$ with $\gamma(0)=\gamma(1)=c(0)=c(1)$ and for some $n\ge 2$.

My question is : rigorously prove that simple closed curves are primitive .

It is visually pretty clear , but I have difficulty proving it. Thanks !

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First, you need to assume that c is essential, so that it doesn't represent the trivial element of \pi_1. Second, you need to be careful, since it's not true on non-orientable surfaces (since the boundary of a Mobius strip is twice the core curve). But once you're orientable, the first homology (abelianization) of the fundamental group is free abelian, and c can be chosen to represent a basis element of this free abelian group. Thus it is primitive in homology, and hence in the fundamental group. – Daniel Groves Nov 3 '11 at 15:13
Thanks , I have edited my question. But when you say " $c$ can be chosen to represent a basis element of the first homology group ", do you assume that c is a non-separating simple closed curve in X and use the result that given any two non-separating simple closed curves $c1,c2$ in an oriented surface $X$ , there exists a global homeomorphism of the surface carrying $c1$ onto $c2$, and then taking $c_1$ as $c$ and $c_2$ and generator of $H_1(X)$ ? Is there a similar result for separating closed simple curves ? – Analysis Now Nov 3 '11 at 15:26
There's a proof in Farb and Margalit's "Primer on mapping class groups". – Richard Kent Nov 3 '11 at 15:32
Oh right, I was assuming separating. But as Richard says, there's a proof in the non-separating case in Farb and Margalit's book... – Daniel Groves Nov 3 '11 at 15:33

Suppose that $c = \gamma^n$ in $\pi_1(X)$. Note that, as $\pi_1(X)$ is torsion free and $c$ is assumed to be non-trivial, the element $\gamma$ generates an infinite cyclic subgroup $\langle \gamma \rangle < \pi_1(X)$. Let $A = X^\gamma$ be the cover of $X$ corresponding to the subgroup $\langle \gamma \rangle$. So $\pi_1(A)$ is also infinite cyclic. Since $X$ is orientable, so is $A$. It follows from the classification of surfaces $A$ is a (non-compact) annulus.
Note that $\gamma$ can be lifted to $A$ and this lift, $\gamma'$, is homotopic to the core curve of $A$. Likewise $c$ lifts to a curve $c'$ and we have $c' = (\gamma')^n$ in $\pi_1(A)$. Since $c$ is simple in $X$ the lift $c'$ is simple in $A$. By the intermediate value theorem (sort of!) the only simple curves in $A$ are isotopic to the trivial curve and to the core curve. Thus $n = \pm 1$ and we are done.
Sam Nead : thanks for the helpful answer ! I just have a quick question about the last line : I understand that $c$ is isotopic to the core curve of $X_\gamma$, since it is non-trivial. But then how exactly are we saying that $n=+/−1$ ? Are we considering the winding number of $c'$ and $\gamma'$ and getting that 1=n. (winding number of $\gamma$),which means n=1 ? And what is the version of intermediate value theorem that you are using here , that proves : simple closed curves in the annulus are either trivial or isotopic to the core curve ? – Analysis Now Nov 17 '11 at 17:06