A topological ("closed") knot is an embedding of a circle in $\mathbb{R}^3$. It's possible for a knot to be distinct from the unknot because there are no free ends to move around and untie the knot. An alternative notion of a knot is an embedding of a closed line segment in $\mathbb{R}^2 \times [0,1]$ such that the two boundary points of the line segment get mapped to the two boundary planes of $\mathbb{R}^2 \times [0,1]$ respectively (one end on each plane), and all other points get mapped to the interior of $\mathbb{R}^2 \times [0,1]$. I'll call this an "open knot". It's possible for an open knot to be distinct from the unknot because, although it has two ends that can move around, they are stuck on the boundary planes, and no part of the knot can get around them, so it can't get untied.

My question is whether these two kinds of knot are fundamentally equivalent. ~~Clearly there are some differences related to chirality/reversibility issues, for example left- and right-handed "overhand" open knots are distinct even though the closed trefoil knot is equivalent to its mirror image (just turn it over). I want to know if that's the full extent of the difference.~~ Whoops, don't know what I was thinking here. If you turn a right-handed trefoil over it remains right-handed.

There's an obvious function mapping from open knots to closed knots - just connect the two ends together. But if you try to invert this function you have to make a choice of where to cut the knot, and you might get different results depending on where you cut it.

Do all choices of where to cut a closed knot result in the same open knot? If not, what's an example?