I'm trying to writing down a proof for the following claim about Dehn-Twists:
Let $\{a_1,...,a_m\}$ be a collection of distinct nontrivial isotopy classes of simple closed curves in a surface $S$ , and assume that $$i(a_i , a_j ) = 0$$ for al $i,j$. Let $\{b_1,...,b_n\}$ be another such collection. Let $p_i,q_i \in \mathbb{Z}\setminus\{0\}$. If $$ T^{p_1}_{a_1}T^{p_2}_{a_2} \cdots T^{p_m}_{a_m} = T^{q_1}_{b_1}T^{q_2}_{b_2}\cdots T^{q_n}_{b_n} $$ in $\text{Mod}(S)$ then $m = n$ and the sets $\{T^{p_i}_{a_i}\}$ and $\{T^{q_i}_{b_i}\}$ are equal.
Which is Lemma 3.17 of Margalit-Farb book. The proof is said to be a straightforward generalisation of the proof given to show $T_a=T_b \Leftrightarrow a =b$
My Attempt I'm able to prove the case where all the exponents are either positive or all negative (not mixed), because I can use use proposition $3.4$ (page $74$) which states, that in this setting, with exponents of the same signs $$ i(M(b),b) = \sum_{i=1}^m|p_i|i(a_i,b)^2$$ where $M$ is my multi-twist. Using Change of Coordinate Principle together with induction on $m$, I'm able to find a suitable simple closed curve that makes the trick (generates an absurd with the above formula applied to the two equal multi-twists) and hence I can prove the claim (details are somewhat boring, I split in two subcases, but this is the rough idea).
The Problem I cannot find a way to deal with the case of mixed sign coefficients. It should be easy but without the proposition $3.4$ I don't know how to control the behaviour of my multi-twist. (There is a slightly generalised version of such proof on Ivanov book, but it only gives a lower bound and an upper bound, and not -according to me- enough control). I tried another approach, but with the change of coordinate principle I cannot find a simple closed curve disjoint from BOTH families of curves, except of one element, because I don't know if the two families intersect with each other. What is disturbing me is that according to Farb-Margalit this should be an easy proof, but I'm having an hard time proving it. Am I missing something?
Such a strong result as Proposition 3.4 is not necessary to do this problem. Perhaps you could get by with something much weaker. Here's a few suggestions.
For instance, suppose you could prove a very weak version of Proposition 3.4 which says the following: for each $b$, if there exists $a_i$ such that $i(a_i,b) \ne 0$, then $i(M_a(b),b) \ne 0$. Here I use $M_a$ to represent the multitwist on the left hand side of your equation. Then could deduce that if $i(a_i,b_j) \ne 0$ for some $i,j$ then $M_a \ne M_b$, because obviously $M_b(b_j)=b_j$ and so $i(M_b(b_j),b_j)) = 0$.
Now, this "weak version" I wrote is not true: for a counterexample, take $b_j$ to have intersection number $1$ with $a_1$ and $a_2$, and intersection number zero with all other $a$'s, and take $p_1=1$, $p_2=-1$. However, I think this is a problem only for small exponents like $p_1=1$ and $p_2=-1$, perhaps the "weak version" is true assuming some lower bound on $|p_i|$. That would still be useful, because the equation $M_a=M_b$, which written in full is $$ T^{p_1}_{a_1}T^{p_2}_{a_2} \cdots T^{p_m}_{a_m} = T^{q_1}_{b_1}T^{q_2}_{b_2}\cdots T^{q_n}_{b_n} $$ implies the equation $M^k_a = M^k_b$, which when written in full becomes $$ T^{k p_1}_{a_1}T^{k p_2}_{a_2} \cdots T^{k p_m}_{a_m} = T^{k q_1}_{b_1}T^{k q_2}_{b_2}\cdots T^{k q_n}_{b_n} $$ because all the individual $T_a$'s commute with each other, and the same for the $T_b$'s.