right side of $R$; ignore the portion of every path of $I_M$ that lies to the right of $\ell$ (and hence outside of $R$). Next, convert $R$ into a polygon $P$ by slightly modifying $R$ at the intersection point of each path in $I_M$ with the right side of $R$ such that the path ends at a vertex of the polygon $P$; see Figure \[fig:ourOuterstring\] for an illustration. First, notice that by definition of $\ell$, every path in $I_M\cap R$ intersects the right side of $R$ in exactly one point. Moreover, since all paths in $I_M$ are [$\llcorner$]{}-type paths, no two paths in $I_M$ would intersect each other on the right of $\ell$ without first intersecting each other inside $R$; hence, $R$ captures all the adjacencies between the paths in $I_M$. The resulting representation models our outerstring graph.

Clearly, $P$ is a simple polygon with $O(N)$ vertices (and, hence $O(N)$ edges). Moreover, each path inside $P$ consists of two segments and so the total number of segments to represent all the paths in $I_M$ is $O(N)$. Therefore, $G_{I_M}$ is outerstring and the above construction is an $O(N)$ geometric representation of $G_{I_M}$.

![A illustration in support of the proof of Lemma 