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A knot is a $S^1$ embedded into $S^3$. Knots $K_0, K_1$ are concordant if there is a locally flat cylinder $C \cong S^1 \times [0,1]$ embedded in $S^3 \times [0,1]$ such that the ends $S^1 \times \{i\}$ are embedded in $S^3 \times \{i\}$ as $K_i \times \{i\}$. This defines a equivalence relation.

The operation of connected sum $J\#K$ makes the set of equivalence classes into an abelian group.

I am stuck on showing how the operation is well-defined, i.e. if $J_0$ is concordant to $J_1$, $K_0$ is concordant to $K_1$, then $J_0\#K_0$ is concordant to $J_1 \# K_1$.

Rolfsen's 'Knots and Links' provides a hint: any concordance may be assumed to be straight on an arc, i.e. after an orientation preserving homeomorphism of $S^3 \times [0,1]$, there is an arc $A\subset S^1$ such that the subset $A \times [0,1]$ of $C \cong S^1 \times [0,1]$ is embedded in $S^3 \times [0,1]$ as the product of an inclusion of $A \subset S^3$ and the identity on $[0,1]$. It is clear that this hint implies what I want to show, but I cannot see why this hint is true.

Any help is appreciated!

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    Someone working in concordance theory told me they didn't think this was possible without Freedman. I believe that all of Freedman's constructions of cobordisms satisfy Rolfsen's hint, but I don't know off hand how to do it with Freedman.. There's might be some neighborhood theorems in the Freedman and Quinn book that might kill this. I imagine there is at least some non-obvious early 20th century decomposition space theory or some Moore-Bing topology that will prove the hint, but I am at least convinced that Rolfsen likely didn't intend the reader to prove it. – 2017-01-25
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    It's certainly provable in the smooth setting. Would you be satisfied with such a proof? – 2017-01-26
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    @MikeMiller Yes, I would. Although Rolfsen probably intended the setting to be PL, I would like an idea of how this can be done. – 2017-01-26

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In the locally flat, PL, and smooth settings, we have the isotopy extension theorem: given an isotopy of a submanifold, it can be extended to an isotopy of the ambient manifold; and if the isotopy is fixed on some submanifold, then the ambient isotopy can be chosen to be fixed there as well. Start by picking a basepoint in $S^1$ and considering the embedded arc in $S^3 \times [0,1]$. It is straightforward to see that there is only one isotopy class of arc in $S^3 \times [0,1]$ in any setting, so we may take the concordance to be constant on this arc.

The part that makes locally flat hard is that we use the local structure of an embedding in the PL and smooth settings: in the former case, triangulate $S^1 \times [0,1]$ so that the embedding is simplicial, and then modify the embedding locally to the arc to make it straight. In the smooth setting, you do the same but with the derivative: you can make it so that the tangent plane of the concordance is constant along the arc, and then (locally to the arc) you take a limit of $f_t(x) = f(xt)/t$ as $t \to 0$. One needs to be somewhat careful in the local construction; ultimately you'll be making it straight on a small interval at a time.

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    Can you please explain why there is only one isotopy class of arc in $S^3 \times [0,1]$? – 2017-01-28
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    @Tsang First, isotope it to lie inside $\Bbb R^3 \times [0,1]$, and to be in general position with respect to a projection onto $\Bbb R^2$. Then the point is, more or less, that any knot in $\Bbb R^3$ is trivial in $\Bbb R^4$, and the way you prove this is by using the extra dimension to change crossings as necessary. – 2017-01-28
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    Yes, I understand it now. Thanks! – 2017-01-28