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Let $K_1,K_2$ be two knots in $S^3$ and assume that there exists a cobordism $(W;K_1,K_2)$ which is invertible from both ends. Does this imply that $K_1, K_2$ are equivalent? In the paper by D.W. Sumners "Invertible Knot Cobordisms", Com. Math. Helv. (46), no.1 (1975), 240-256, he claims that in such a case $W$ is a $h$-cobordism between $S^3\setminus K_1$ and $S^3\setminus K_2$. In particular, the fundamental groups of their complements are isomorphic. The result by Sumners is stated in a more general situation for $n$-knots in $S^{n+2}$, so probably when $n=1$ something more can be deduced.

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  • $\begingroup$ Sorry, the $h$-cobordim between $S^3\setminus K_1$ and $S^3\setminus K_2$ is not $W$, but $S^3\times I\setminus W$. $\endgroup$ Oct 6, 2015 at 21:51
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    $\begingroup$ You should edit the question to make these changes. $\endgroup$ Oct 6, 2015 at 21:59
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    $\begingroup$ What does "invertible" mean in this context? $\endgroup$ Oct 6, 2015 at 22:14

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Yes, $K_1$ and $K_2$ are equivalent. It is known that knots $K_1$ and $K_2$ are equivalent if and only if there's an isomorphism between the fundamental group of their complements that preserves the peripheral structure (ie the meridian and longitude).

By assumption, there is a cobordism $C_1$ from $K_1$ to $K_2$ and a cobordism $C_1'$ from $K_2$ to $K_1$ such that the composition $C_1 \cup_{K_2} C_1'$ is a product cobordism. This is an invertible cobordism from $K_1$ to $K_2$. Sumners shows that if there is also an invertible cobordism $C_2$ from $K_2$ to $K_1$, then the complement of any of these cobordisms is a relative h-cobordism between the knot complements. In particular, the fundamental groups of the complements are isomorphic, and include isomorphically into the fundamental group of the complement of the cobordism. Since this isomorphism is induced by inclusions, it's easy to check that it respects meridians and longitudes, and hence the knots are equivalent.

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  • $\begingroup$ Thanks a lot, I didn't know the result that if the isomorphism between the fundamental groups preserves the peripheral structure, then the knots are equivalent. I think this solves our question. $\endgroup$ Oct 7, 2015 at 12:43
  • $\begingroup$ @Juan: You're quite welcome. Subsequent to this discussion, I posted a preprint with Auckly, Kim, and Melvin that shows that for closed 3-manifolds the invertible cobordism relation is asymmetric. The proof uses some properties of 3-manifold groups that follow from geometrization. We are revising the paper now and I might put in the additional observation above about respecting peripheral structure. $\endgroup$ Jan 27, 2020 at 14:53

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