Reduction of self-intersections without reducing the geometric intersection Let $F$ be a hyperbolic surface. Given a closed curve $a$, let $\bar{a}$ denotes the free homotopy class of $a$. Let $i(\bar{a},\bar{b})$ denotes the geometric intersection number and $i(\bar{a})$ denotes the self intersection number. 
Suppose $a$ and $b$ are two given closed curves in $F$ with the following properties:
1) $a$ is a simple closed curve.
2) $i(\bar{b})=k\neq 0$.
3) $i(\bar{a},\bar{b})=m\neq 0$.
Q) Does there exists a cover $\widetilde{F}$ of $F$ such that:
1) $a$ and $b$ lifts to closed curves $\alpha$ and $\beta$ respectively in $\widetilde{F}$.
2) $i(\bar{\beta})<k$.
3) $i(\bar{\alpha},\bar{\beta})=m$.
In other words, can we lift a pair of curves (one of which is simple) to a cover without reducing the geometric intersection number but reducing the self intersection number of one of them. 
PS 1: I have tried LERF property and double coset separabilty of surface groups, but was unable to prove it.
PS 1: Any reference of any kind will be helpful. Thanks in advance.
 A: I don't think this is possible in general. The point is
that if $i(\alpha, \beta)= m$, then all the arcs of $b-a$
lift uniquely to $\tilde{F}$ to arcs of $\beta-\alpha$. 
In fact, the graph consisting
of the union of $a$ and $b$ along their points of intersection
lifts from $F$ to $\tilde{F}$. One may then close these
arcs to closed (basepointed) loops which also must lift. But choosing
$b$ complicated enough, one may ensure that these loops
generate the fundamental group, so no cover is possible. I
give an example in the figure, where the closed-off arcs
of $b$ give standard generators for the fundamental group
of the surface. 

Maybe it's easier to understand the case that $b$ is embedded.
Consider the graph $a\cup b$ in the following diagram:

If there is a cover $\tilde{F}\to F$ that $a$ and $b$ lift to so that the
intersection number is the same, then the entire graph 
$a\cup b$ lifts. But $\pi_1(a\cup b)\twoheadrightarrow \pi_1(F)$
surjects, so $\pi_1(\tilde{F})\twoheadrightarrow \pi_1(F)$
surjects, i.e. is a trivial (connected) cover. 
