The basic arcs detect the identity braid

A preliminary identification. Suppose $g\in G$ satisfies $g(b_k)\simeq b_k$ for all $k$. Then $[g]=1$ by [L2], and in particular $g(b_k)$ is isotopic to $b_k$ for every $k$ and the intersection table of $g$ with the basic set is the identity table.

The intersection table determines the action on the basic curves. Let $c$ be admissible and suppose $I(b_j,c)=I(b_j,b_k)$ for all $j$. For $k<m$, [L3] gives $c\simeq {\tau }_k^{\nu }(b_k)$ for an integer $\nu \in \{-1,0,1\}$, with $\nu \le 0$ when $k=0$; for $k=m$, it gives $c\simeq b_m$ directly. Consequently, applying this to $c=f(b_k)$ for each $k$ separately, there are integers ${\nu }_0\in \{-1,0\}$ and ${\nu }_1,\dots ,{\nu }_{m-1}\in \{-1,0,1\}$ with $f(b_k)\simeq {\tau }_k^{{\nu }_k}(b_k)(0\le k\le m-1),f(b_m)\simeq b_m\mathrm{.}$

An explicit model for $f$. Let $g:={\tau }_0^{{\nu }_0}{\tau }_1^{{\nu }_1}\cdots {\tau }_{m-1}^{{\nu }_{m-1}}\in G$. Since the twists commute and ${\tau }_j$ fixes the complementary basic arcs by [L4], one has $g(b_k)\simeq {\tau }_k^{{\nu }_k}(b_k)$ for $k\le m-1$ and $g(b_m)\simeq b_m$; comparing with step 1.2, $f(b_k)\simeq g(b_k)$ for every $k$. Applying step 1.1 to $g^{-1}f$, which fixes each $b_k$ up to isotopy, gives $[f]=[g]$.

The same for the square, and comparison of exponents. Applying the same argument to $f^2$, whose intersection table with the basic set is also the identity table by hypothesis, produces integers ${\mu }_0\in \{-1,0\}$ and ${\mu }_1,\dots ,{\mu }_{m-1}\in \{-1,0,1\}$ with $f^2\simeq {\tau }_0^{{\mu }_0}\cdots {\tau }_{m-1}^{{\mu }_{m-1}}\mathrm{.}$ Since $[f]=[g]$, one has $[f^2]=[g{]}^2={\prod }_j{\tau }_j^{2{\nu }_j}$, and therefore ${\prod }_{j=0}^{m-1}{\tau }_j^{2{\nu }_j-{\mu }_j}\simeq 1.$

Freeness kills the exponents. By the freeness of the twist subgroup [L4], the relation ${\prod }_j{\tau }_j^{2{\nu }_j-{\mu }_j}\simeq 1$ forces $2{\nu }_j={\mu }_j$ for every $j$. Since each ${\mu }_j$ lies in $\{-1,0,1\}$ and each $2{\nu }_j$ is even, the only possibility is ${\nu }_j={\mu }_j=0$ for all $j$; hence $[f]=[g]=1$, as required.

Conclusion. The identity intersection table for $f$ and $f^2$ forces $[f]=1$. AC is inherited through the braid/mapping-class dictionary and the supplied well-definedness and isotopy invariance of geometric intersection numbers used in [L1] and [L3]; the local counts and exponent bookkeeping are finite.