Given two links $K = K_1 \cup \dotsb \cup K_n$ and $L = L_1 \cup \dotsb \cup L_n$, where each $K_i$ and $L_j$ are oriented knots, can we define the connected sum $K\#L$ by taking the connected sum of corresponding components, such that $$ K\#L = (K_1\#L_1) \cup \dotsb \cup (K_n\#L_n)? $$ Is the result well-defined? If it is, can I have a reference to this definition?
2026-03-26 09:29:49.1774517389
Component-wise connected sum of links
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Even if you order the link components, this operation is not well-defined up to isotopy. Here's a counterexample: Letting $K=K_1 \cup K_2$ be the two-component unlink and $L=L_1\cup L_2$ the Hopf link, the "obvious" way of defining $K \# L$ yields the Hopf link $L$ again. However, if we choose the unlink $K$ to have a more complicated projection, the resulting link need not be the Hopf link. Check out the figure below. According to SnapPea, the resulting link's complement in $S^3$ admits a hyperbolic structure. The complement of the Hopf link is homeomorphic to $S^1 \times S^1 \times \mathbb{R}$, which is geometrically Euclidean. Therefore the two links are inequivalent.
I should note that some of the more basic invariants are unable to distinguish between inequivalent "connected sums" of links.