A genus-g knot cobordism bounds the tau difference

Choose a smooth embedded arc in $S$ joining its two boundary components. After orientation-preserving isotopies of the endpoint knots inside their product collars, which preserve their knot types, choose a standard reference arc with both endpoints at the same point of $S^3$; its tubular complement is $(S^3\setminus \mathrm{int}{D}^3)\times [0,1]$, a four-ball after rounding corners. Since $S^3\times [0,1]$ is simply connected, the chosen arc is homotopic rel endpoints to this reference. A generic one-parameter homotopy is an isotopy through embedded arcs: the double-point condition has parameter-domain dimension $1+1+1=3<4$, and the nonimmersion condition has dimension $1+1=2<4$. Extend its velocity field from the compact arc family to a time-dependent ambient vector field using a tubular neighborhood and a bump function; integrating this field gives an ambient isotopy carrying the arc neighborhood to the reference neighborhood. Delete that tubular neighborhood from the ambient product and delete the corresponding open rectangular strip from $S$. The remaining surface is connected and has one boundary component, the oriented sum $K_1\mathrm{\#}(-K_0)$ under [F2]. Since $\chi (S)=-2g$, cutting along this arc raises Euler characteristic by one to $1-2g$; the resulting surface has one boundary component, so its genus is $g$.

Apply Hom's imported genus bound [F3] to the surface from step 1.1: $\mathrm{\mid }\tau (K_1\mathrm{\#}(-K_0))\mathrm{\mid }\le g_4^{\mathrm{s}\mathrm{m}}(K_1\mathrm{\#}(-K_0))\le g$. Hom's homomorphism statement [F3] gives $\tau (K_1\mathrm{\#}(-K_0))=\tau (K_1)-\tau (K_0)$, proving the asserted inequality. For a concordance annulus $g=0$, so the endpoint values agree. If one endpoint is the standard unknot, [F4] gives $\tau (U)=0$, and the inequality is the four-ball genus estimate for the other endpoint.