Bridge number one characterizes the unknot

Suppose $b(K)=1$. By [A1] and [F1], the nonempty set of maximum counts has least value $1$, so choose a Morse representative $K^{\mathrm{\prime }}$ with exactly one local maximum. On the circle, nondegenerate critical points alternate between maxima and minima; compactness gives finitely many, and a function on a circle has at least one minimum. Hence $h\mathrm{\mid }{K}^{\mathrm{\prime }}$ has exactly one minimum. Choose a regular value between these two critical values. Its level sphere meets $K^{\mathrm{\prime }}$ in exactly two points, and the upper and lower balls contain one proper arc each, with respectively one maximum and one minimum.

Fix a simple arc $\beta$ on the level sphere joining the two intersection points. Apply [F2] to the upper arc and the lower arc, using this same $\beta$. The resulting bridge disks lie in opposite balls and meet only along $\beta$; their union, with its corner smoothed, is an embedded disk whose boundary is $K^{\mathrm{\prime }}$. Thus $K^{\mathrm{\prime }}$ has genus zero, and [F3] makes it ambient isotopic to the standard unknot.

Conversely, choose an orientation-preserving rotation in $SO(4)$ carrying the fixed coordinate 2-plane of the standard unknot $U$ to the plane spanned by $e_1$ and $e_4$. A path of rotations from the identity gives an ambient isotopy, and on the rotated circle the fixed height is a sine or cosine with exactly one maximum. Hence $b(U)\le 1$. Every Morse function on a compact circle has at least one maximum, so $b(U)\ge 1$ and $b(U)=1$. Ambient-isotopic knots have the same set of representative maximum counts, so every unknot has bridge number one.