The standard nested twists generate a free abelian subgroup

Pair windings detect the twist exponents. For $0\le k<m$, follow the pair $(q_k,q_m)$ during an unmarked isotopy from the identity to ${\tau }_j$, and take the winding of their nonzero difference. In the standard circular model of [L2], if $j\le k$ both points are inside the rotating disk and their difference makes one full turn, giving $1$. If $j>k$, only $q_m$ moves, and its path lies in a disk not containing $q_k$, so the difference loop has winding zero. Concatenation gives pair winding ${\sum }_{j=0}^k{e}_j$ for the standard unmarked isotopy to ${\prod }_j{\tau }_j^{e_j}$; negative exponents reverse the corresponding paths.

A relation has zero pair windings. Suppose that product is isotopic to the identity relative to the boundary and the marked set. Append that marked isotopy to the unmarked isotopy in step 1.1; the marks are constant on the appended part, so it does not change the pair windings. This produces a loop $f_t$ in the group of boundary-fixed homeomorphisms of the unmarked unit disk. It contracts by the explicit Alexander formula $A_s(f)(x)=sf(x\mathrm{/}s)$ for $\mathrm{\mid }x\mathrm{\mid }\le s$, $A_s(f)(x)=x$ for $\mathrm{\mid }x\mathrm{\mid }\ge s$, and $A_0(f)=\mathrm{i}\mathrm{d}$. The two formulas agree on $\mathrm{\mid }x\mathrm{\mid }=s$ because $f$ fixes the boundary; each $A_s(f)$ is a homeomorphism, and continuity at $s=0$ follows from $\mathrm{\mid }{A}_s(f)(x)-x\mathrm{\mid }\le 2s$. Applying this contraction to the loop $f_t$ gives a homotopy of each pair's nonzero difference loop to the constant loop. Thus every winding from step 1.1 is zero.

The triangular equations force all exponents to vanish. For $k=0$ the zero-winding equation is $e_0=0$. For each $1\le k<m$, subtract the equation for $k-1$ from that for $k$ to obtain $e_k=0$. Hence the only relation among the commuting twists is the trivial exponent vector.

Conclusion. Commutation [L1] defines a surjective homomorphism ${\mathbb{Z}}^m$ onto the generated subgroup, and step 3.1 proves it is injective. Thus that subgroup is free abelian of rank $m$, including $m=1$, when the single pair winding detects ${\tau }_0$. The proof uses explicit isotopies and finitely many windings and requires no choice axiom.