# Connected sum of a link and a knot

Let $$L=K_1 \cup K_2$$ be a two-components link in a copy of $$S^3$$ and let $$K$$ be a knot, thought in a different copy of $$S^3$$. In other words, we have two couples $$(S^3, L)$$ and $$(S^3, K)$$. Let us define a notion of "connected sum" between objects of this kind: we choose a component of $$L$$ (for example $$K_1$$) and two arcs $$S \subset K_1$$ and $$S' \subset K$$. We define the connected sum of the couples $$(S^3, L)$$ and $$(S^3, K)$$ along $$K_1$$ as the object $$(S^3 \# S^3 \simeq S^3, K \# K_1 \cup K_2)$$, where the connected sum of the two copies of $$S^3$$ is performed along disks which intersect the links in the chosen intervals $$S$$ and $$S'$$.

I am not sure that this operation is well defined; it would be nice if it was but I wouldn't bet on that. What I would like to prove is that if $$L$$ is a split link, then the result of this operation is again a split link. The problems seems to be the fact that the bars used to perform the connected sum of knots could intersect the other component ($$K_2$$); I think that it could be possible to perform the connected sum of knots "inside" the disks used to perform the connected sum of the two copies of $$S^3$$. Is it correct? Would it be enough to state the the resulting link is split, under the hypothesis that $$L$$ is split? Thanks in advance.

P.S. I found this definition of connected sum of links in "Three-dimensional link theory and Invariants of Plane Curve singularities" by Eisenbud & Neumann. This definition can be in fact easily generalized to the connected sums of two links, but things may turn to be more complicated in that case.

• It seems that I have only to check that this construction does not depend on the diagram (but this should not be difficult, using Reidmeister moves I can always obtain a split diagram) and that I can take $K$ to be on one side of the line...for this second part can I use an argument like "all disks are isotopic"? So that I can choose one such disk in $S^3$ so that the projection stays on one side of the line? I hope I was clear. Feb 18, 2021 at 15:29