Smooth and locally flat topological concordance are equivalence relations

Reflexivity is witnessed by the product annulus $K_0\times I\subset S^3\times I$, oriented so its boundary is $-K_0\times \{0\}\bigsqcup K_0\times \{1\}$. If a representative is changed by an ambient isotopy $H_t$, reparameterize time so $H_t$ is constant near $t=0$ and $t=1$. Its trace $(x,t)\mapsto (H_t(x),t)$ then has product endpoint collars and is a concordance annulus; it is smooth in the smooth case and locally flat in the topological case because it is itself smooth. For an existing concordance, glue the trace to the matching product collar: use the reversed trace at the lower end and the forward trace at the upper end. The glued surface is an annulus in the same category, so replacing either endpoint by an ambient-isotopic representative preserves concordance. Thus the relation is well-defined on ambient-isotopy classes.

If an annulus $A$ gives a concordance from $K_0$ to $K_1$, apply $(x,t)\mapsto (x,1-t)$ to its image. Reorient the resulting annulus so the boundary at the new lower end is $-K_1$ and the boundary at the new upper end is $K_0$. The map and reorientation preserve smoothness in the smooth category and local flatness and product collars in the topological category. This gives symmetry.

Let $A$ be a concordance from $K_0$ to $K_1$ and $B$ one from $K_1$ to $K_2$ in the same category. Glue the upper product collar of $A$ to the lower product collar of $B$ along their common oriented knot $K_1$. The two collar coordinates combine to a single collar across the gluing hypersurface, so the annuli join without a corner in the smooth category and their local-flat pair charts join in the topological category. Gluing two annuli along one boundary circle gives an annulus; the intermediate boundary orientations $K_1$ and $-K_1$ cancel, leaving boundary $-K_0\bigsqcup K_2$. This gives transitivity.

Steps 1.1–1.3 prove reflexivity, symmetry, transitivity, and representative independence for both relations.