Step 1: Setup. Given an oriented link $L$, we may assume $L$ is in general position with respect to a chosen axis $A$ (a line in $\mathbb{R}^3$). We will isotope $L$ so that it winds monotonically around $A$, making it the closure of a braid.

Step 2: Choose an axis and projection. Consider $\mathbb{R}^3$ with cylindrical coordinates $(r,\theta,z)$ and let $A$ be the $z$-axis. In the projection to the $(r,\theta)$-plane, the link $L$ appears as a closed curve. The angular coordinate $\theta$ may not be monotonically increasing along $L$ -- there may be arcs where $\theta$ decreases (the link "goes backwards" around the axis).

Step 3: Eliminate bad arcs. A bad arc is a subarc of $L$ along which $\theta$ decreases. For each bad arc $\alpha$:

• Choose a point $P$ far from $L$ on the axis $A$ (far above or below the projection).
• Replace $\alpha$ by a detour: isotope $\alpha$ by swinging it around the axis $A$ in the positive direction, passing near $P$, so that $\theta$ is now increasing along the modified arc.

More precisely, if $\alpha$ goes from angle $\theta_1$ to $\theta_0 < \theta_1$, we replace it by an arc that increases angle by $2\pi - (\theta_1 - \theta_0)$, achieving the same net change while always moving in the positive $\theta$ direction. The point $P$ is chosen so that the detour avoids all other strands of $L$.

Step 4: Result is a braid closure. After eliminating all bad arcs, the modified link $L'$ (isotopic to $L$) has $\theta$ monotonically increasing along every component. Cutting $L'$ along a half-plane $\theta = \theta_0$ yields a braid $\beta \in B_n$ (where $n$ is the number of intersection points with the half-plane), and $L' = \hat{\beta}$.

Step 5: Finiteness. There are only finitely many bad arcs (since $L$ is a finite union of smooth arcs in general position), so the procedure terminates. $\square$