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In "4-manifolds and Kirby Calculus" by Gompf and Stipsicz, there is a nice description of how to get the Kirby diagram of $S^1 \times M^3$, given a Kirby diagram of $M^3$. Basically, one thickens the diagram, adds one 1-handle and connects it to the existing 1-handles with 2-handles in a specific way. Is there a similar algorithm for surgery diagrams?

I learned that there is relative Kirby calculus (Are there Kirby diagrams for manifolds with boundaries?) and it is very easy to express $I \times M^3$ given a surgery diagram for $M^3$, but I don't know how to express gluing along a boundary in this language, more specifically, gluing ${0} \times M^3$ to ${1} \times M^3$ to yield $S^1 \times M^3$.

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