A genus-one spanning surface for the unknot is not minimal

Let $U$ be the standard round unknot and $D$ a smooth embedded spanning disk. Remove the interiors of two disjoint small disks from $\mathrm{Int}D$ to obtain a pair of pants $P$. Join the two new boundary circles by a simple arc in a small ball on one side of $D$ and disjoint from $U$; the lateral boundary of a sufficiently thin tubular neighborhood of that arc is an embedded annulus $A$ meeting $P$ only in its two attaching circles. Glue with matching orientations. Then $F=P\cup A$ is connected, compact, smooth, embedded, oriented, and has only boundary $U$. Choose finite cell structures with both attaching circles subcomplexes. Cell-count inclusion-exclusion gives $\chi (P)=\chi (D)-2\chi (D^2)+2\chi (S^1)=1-2+0=-1$; then $\chi (F)=\chi (P)+\chi (A)-2\chi (S^1)=-1+0-0=-1$.

The surface $F$ has $b=1$. By [F3] and [A1], $-1=\chi (F)=2-2g(F)-1$, hence $g(F)=1$. The original disk $D$ has $b=1$ and $\chi (D)=1$, so [F3] gives $g(D)=0$.

By [F1], both $D$ and $F$ are admissible spanning surfaces for $U$. Therefore [F2] and nonnegativity of surface genus give $g(U)=0$, while step 2.1 gives $g(F)=1$. Thus $F$ is not minimum genus and the refuted statement is false.