Knot genus is additive under connected sum

Remove a small boundary half-disk from each minimum genus surface $F_K,F_J$ in the chosen connected-sum balls and glue their exposed boundary intervals. This is an embedded oriented boundary connected sum with boundary $K\mathrm{\#}J$. Attaching the single connecting band changes Euler characteristic by $-1$, so $\chi (F)=\chi (F_K)+\chi (F_J)-1$. By [F2], $\chi (F_K)=1-2g(K)$ and $\chi (F_J)=1-2g(J)$; hence $\chi (F)=1-2(g(K)+g(J))$ and $g(K\mathrm{\#}J)\le g(K)+g(J)$.

Choose a minimum genus surface $F$ for $K\mathrm{\#}J$ and, among such surfaces transverse to the standard sum sphere $S$, one minimizing the finite number of circular components of $F\cap S$. Such a surface exists by [F1], [F3], and well-ordering of the nonnegative integers. The intersection is a compact one-manifold whose boundary is the two points $(K\mathrm{\#}J)\cap S$, hence consists of one arc $\alpha$ and finitely many circles.

If circles occur, choose one innermost on $S$. Since the intersection circles are disjoint and the arc $\alpha$ is connected and misses them, finite nesting gives a disk side of some circle containing neither $\alpha$ nor another intersection component. By [F4] this is an embedded disk $D\subset S$, disjoint from $K$.

If the selected circle $c$ is nonseparating on $F$, cut $F$ along $c$ and cap the two new boundary copies by parallel copies of $D$, rounded in a product neighbourhood of $c$. The caps are disjoint and the boundary remains the original knot. Cutting along a circle preserves Euler characteristic and the two caps add $2$, so by [F2] the new genus is $g(F)-1$, contradicting minimality.

If $c$ separates $F$, cutting gives a component $F_K$ containing the knot boundary and a component $F_0$ with boundary $c$. Let their genera be $h_K,h_0$. Since $F_K$ has two boundary components and $F_0$ has one, Euler characteristic additivity along the cut and [F2] give $g(F)=h_K+h_0$. Cap $F_K$ by a parallel copy of $D$ and discard $F_0$. If $h_0>0$ this is a lower genus spanning surface, contradiction; if $h_0=0$, it is another minimum genus surface with fewer intersection circles, contradicting the choice in step 1.2. The cap is embedded after rounding in the local product neighbourhood from step 2.1. Thus no circle can occur.

With no circles, $F\cap S=\alpha$. Since $S$ separates the two knot pieces in the standard sum model, the proper arc $\alpha$ separates $F$ into two connected surfaces $F_1,F_2$, each with one boundary component. Cutting along an arc increases total Euler characteristic by $1$, so [F2] gives $g(F)=g(F_1)+g(F_2)$. The two surfaces are connected because any component of either side not incident to $\alpha$ would already be a separate component of $F$; the sole arc has just one collar on each side. Their boundary circles consist of the corresponding knot subarc and $\alpha \subset S$, and hence represent $K$ and $J$ by [L1] and the original connected-sum datum. Thus $g(K)+g(J)\le g(F_1)+g(F_2)=g(F)=g(K\mathrm{\#}J)$. Together with step 1.1 this proves the full equality.