The right trefoil has genus one

Use the standard right-trefoil diagram, the closure of the positive two-strand braid ${\sigma }_1^3$, with both braid strands directed downward and the closing arcs directed back to the top. It has exactly three crossings. At each crossing the oriented smoothing pairs each upper endpoint with the lower endpoint in the same vertical position, rather than the diagonal pairing. Smoothing all three crossings therefore leaves two vertical strands; each joins its own closing arc to form one circle. Thus $s=2$ and $c=3$. The unsmoothed braid has the transposition permutation (three interchanges), so its closure has one knot component. The algorithm attaches three half-twisted bands between the two disk records, with twist directions restoring these three positive crossings. By [F1] and [A2] the result is a connected compact smooth oriented embedded spanning surface for $T$, and its genus is $(1-2+3)\mathrm{/}2=1$ (equivalently $\chi =2-3=-1$). Hence the nonnegative minimum in [F5] satisfies $g(T)\le 1$.

If $g(T)=0$, a minimizing surface exists by [F5], and [F3] makes it a smooth disk $D\subset S^3$. Let $c:S^3\times [0,ϵ)\to B^4$ be a collar from [F4]. A boundary collar of $D$ gives a smooth function $h:D\to [0,ϵ)$ that equals the inward collar coordinate near $\mathrm{\partial }D$, is positive on $\mathrm{Int}D$, and vanishes exactly on $\mathrm{\partial }D$. The map $x\mapsto c(x,h(x))$ is injective because the collar projection is $x$, and its differential is injective because its first coordinate is the inclusion of $D$; at $\mathrm{\partial }D$, the positive derivative of $h$ in the inward conormal makes the disk proper and transverse to $\mathrm{\partial }{B}^4$. Thus it is a smooth proper disk in $B^4$ with boundary $T$, contradicting [F2]. Hence $g(T)\mathrm{\ne }0$.

A nonnegative integer at most $1$ and not equal to $0$ is $1$. Hence $g(T)=1$.