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It is a well-known fact that $S^2 \times S^1$ can be obtained by $0$-surgery on unknot.

What about the $(-1)$-surgery on $S^2 \times S^1$? It seems the resulting manifold, say $W$, bounds contractible manifold.

But I cannot prove it yet or refutes my argument. Any help will be appreciated.

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This is true, with some points to clarify. First, you are presumably talking about surgery along a knot that generates the first homology (and hence fundamental group) of $S^1\times S^2$. Then the result of adding the corresponding 2-handle to $S^1\times B^3$ is contractible. The construction (called a `Mazur manifold') goes back to Mazur's paper, A Note on Some Contractible 4-Manifolds, Annals 1961.

The other point is that framing as an integer is not a priori defined for a knot that represents a homology class of infinite order. But fortunately the statement is true for any framing (as in choice of trivialization of the normal bundle). I'd suggest some basic reading about 4-dimensional handle calculus, as in the book of Gompf-Stipsicz.

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  • $\begingroup$ Thanks for the reference. To clarify the notion, let me ask you one more question. If we replace $S^2 \times S^1$ by $S^3$ and perform $−1$-surgery on $S^3$ (following your description), may we obtain a manifold bounding rational homology ball? $\endgroup$
    – M. Alessandro Ferrari
    Commented Apr 20, 2019 at 10:03
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    $\begingroup$ Not necessarily; it depends on the knot along which you do that surgery. I actually think these questions are better suited to Mathstackexchange than to this forum. $\endgroup$
    – Danny Ruberman
    Commented Apr 20, 2019 at 13:36

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