Let $\Sigma$ be a compact oriented connected bordered surface other than the pair of pants. Let $\Gamma:=\{\gamma\_i\}$ be a finite collection of simple closed curves on $\Sigma$ such that each component of $\Sigma\setminus\cup\_i\gamma\_i$ is homeomorphic to either $\Bbb S^1\times (0,1)$ or $\{z\in \Bbb R^2:|z|<1\}$. Notice that $\gamma\_i$ may intersect with $\gamma\_j$.



> 
> Suppose $c$ is a closed curve (*not necessarily simple*) on $\Sigma$
> such that $\text{GI}(c,\gamma\_i)=0$ for each $i$, where $\text{GI}$
> denotes the geometric intersection number. Is it true that $c$ can be
> freely homotoped so that $c\cap \gamma\_i=\varnothing$ for each $i$?
> 
> 
> 



A closed curve can be homotopic to remove all intersections with a filling $\Gamma$ if it has zero geometric intersection numbers with $\Gamma$
This is true, here's a proof, by a kind of "Whitney trick".


Perturb the set of curves $\{\gamma\_i\} \cup \{c\}$ to put it into general position, so they are pairwise transverse and there is no triple point. Let $|c| = \sum\_i |c \cap \gamma\_i|$.


For each $i$ such that $|c \cap \gamma\_i| > 1$, since $GI(c,\gamma\_i)=1$ it follows that $\Sigma - (c \cup \gamma\_i)$ contains a component whose closure $B$ is a *bigon of $c$ and $\gamma\_i$*, meaning a closed disc having the property that $\partial B = \alpha \cup \beta$ where $\alpha = B \cap c = \partial B \cap c$ is a subarc of $\partial B$ and $\beta = B \cap \gamma\_i = \partial B \cap \gamma\_i$ is a subarc of $\partial B$.


If $c$ is not already disjoint from the $\gamma\_i$'s then, as $i$ varies and $B$ varies over all bigons of $c$ and $\gamma\_i$, there exists a bigon $B$ that is innermost with respect to inclusion. Letting $B$ be a bigon of $c$ and $\gamma\_i$, it follows that $B \cap \bigcup\_{j \ne i} \gamma\_j$ is a union of arcs in the $\gamma\_j$'s that cross from $\alpha$ to $\beta$, each such arc having one endpoint on $\alpha$ and the other endpoint on $\beta$. Now isotope $c$ to push $\alpha$ across $B$ and out past $\beta$ on the other side of $B$. This reduces $|c|$ by $2$.


By induction, $|c|$ can be reduced to $0$, at which point $c$ is disjoint from each of the $\gamma\_i$'s.


