Are isotopic transversal curves on a foliated surface transversally isotopic? Let $F$ be an orientable surface (possibly with boundary) with a foliation $\cal F$ with $k$-prong saddle singularities only for $k\geq 3,$ (as in figure borrowed from Farb-Margalit book). Suppose that $\alpha, \beta$ are transversal multi-curves which are isotopic in $F$. Are they necessarily transversally isotopic? No, if $\alpha, \beta$ are separated by a Reeb component. But what about a Reebless foliation? If still not, is there a broad, natural condition on $\cal F$ which guarantees that to be true?
If we allow for additional singularities in $\partial F$, then we need to assume that the endpoints of $\alpha$ and $\beta$ coincide as a transversal isotopy cannot move an endpoint through a singularity.
 
 A: "Reebless" is not strong enough for this to be true. One should also assume that there is no "half-Reeb" annulus, i.e. for every circle leaf $C$ and for each side of $C$ the nearby leaves on that side are not allowed to spiral into $C$.
With that assumption, let me describe the proof for the closed surface case; it is easily adapted to cover the bounded case.
By perturbing $\alpha,\beta$ through transversal curves, we may assume that $\alpha,\beta$ are transverse to each other. Their intersection $\alpha \cap \beta$ consists of $n$ transverse intersection points for some $n \ge 0$. We proceed by induction.
If $n=0$ then $\alpha\cup\beta$ bounds an annulus $A$ with a singular foliation transverse to the boundary. By the index formula, the foliation restricted to $A$ has no singularities. Also, there cannot be any circle leaf in $A$, because such a circle would be isotopic to a core curve of $A$, and it would follow that $A$ contains a half-Reeb annulus. Thus $A$ is homeomorphic to $S^1 \times [0,1]$ with the foliation into arcs $x \times [0,1]$. Now we can transversely isotope $\alpha$ to $\beta$ through $A$.
If $n > 1$ then there exist arcs $a \subset \alpha$, $b \subset \beta$ such that $a \cap b = \partial a \cap \partial b = \{x,y\}$, and the simple closed curve $a \cup b$ bounds a disc $D \subset S$ whose interior is disjoint from $\alpha \cup \beta$. Furthermore, at the two endpoints $x,y$ the foliation is externally tangent to $D$. By the index formula, the foliation restricted to $D$ has no singularity, and has neither a closed leaf nor a leaf which has both endpoints on $a$ or on $b$. It follows that $D$ is foliated like the vertical arcs in $\mathbb{R}^2$ restricted to 
$$\{(x,y) \mid x^2 + (y-1)^2 \le 4, x^2 + (y+1)^2 \le 4\}
$$
Now we can transversely isotope $\alpha$ across $D$ so as to remove two points of intersection.
