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!
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.