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For $f:S\to M$ a knot in a 3-manifold, we can construct a 3-manifold $N$ by a $0/1$-type Dehn surgery along $f$:

  1. First remove from $M$ a solid torus which is a tubular neighbourhood of the knot $f$;
  2. Thereafter $N$ is the result of sewing this solid torus back in $M$ such that the meridian disc goes once time along the longitude and no times along the meridian.

Does the integral cohomology groups $H^1(N;{\mathbb{Z}})$ and $H^2(N;{\mathbb{Z}})$ depend only on the integral cohomology of $M$ or do they also depend on how the knot $f$ sits in $M$?

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1 Answer 1

They depend on how $f$ sits inside $M$. Suppose $M=S^2\times S^1$. Then if a knot is isotopic to $\{x\}\times S^1$, $0$-framed Dehn filling gives back $S^2\times S^1$. On the other hand, if the knot lies in $S^2\times \{y\}$, then Dehn filling gives $S^2\times S^1 \# S^2\times S^1$.

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