Why a special ball in $S^3$ is unique? I'm studying : An introduction to knot theory(by: W.B. Raymond Lickorish). 
To prove composition of two oriented knots is unique, Lickorish has written:"regarded $K_1$ and $K_2$ as being in distinct copy of $S^3$, remove from each $S^3$ a ball that meets the given knot in an unknotted spanning arc" then he have said: "Some basic piecewise linear theory shows that ball meeting the knot in unknotted spanning arcs are essentially unique" 
I want to know how I can prove the second proposition. Is there proof of this result in any article or book? 
 A: You can pick up some of this theory from the likes of Hatcher's notes on 3-manifolds or Hempel's or Jaco's books on 3-manifold topology, or perhaps even Rolfsen's book on knot theory.  Lickorish sprinkles some here and there at various points in his book.
The intuition for a ball $B\subset S^3$ meeting $K$ as a trivial ball-arc pair being "essentially unique" is that you can shrink $B$ down so that it can be thought of as a tiny bead on $K$.  Then, any two beads on $K$ are equivalent by sliding them around.
A trivial ball-arc pair $(B,\alpha)$ is a "ball $B$ with an unknotted spanning arc $\alpha\subset B$", which more precisely is a pair that is, as a subspace, PL homeomorphic to the pair $$[0,1]\times (D^2,0)=([0,1]\times D^2,[0,1]\times 0).$$
Let $(B,\alpha)\subset (S^3,K)$ be a trivial ball-arc pair.
By Alexander's theorem (that for a PL embedded $2$-sphere $S\subset S^3$, the closure of each component of $S^3-S$ is PL homeomorphic to a $3$-ball with boundary $S$) and Alexander's trick, we can get an isotopy of $S^3$ that isotopes $B$ to a standard $3$-ball in $S^3$ while isotoping $\alpha$ to a diameter of $B$.  Alexander's trick uses the PL homeomorphism that characterizes a trivial ball-arc pair.
Once in this position, we can make a knot diagram for $K$ where $\alpha$ is indicated by small arc in the diagram away from any crossings, where $B$ projects onto a small disk with the arc as its diameter.  We are allowed to slide this arc around the diagram (crossing crossings) since there is a corresponding isotopy of $S^3$ to implement this move.  Reidemeister moves (which correspond to certain isotopies of $S^3$) are allowed at least if the arc is not in the way, but the arc can always be moved to be out of the way.  Also, isotopies of the diagram itself correspond to isotopies of $S^3$ as usual.  From here, it should be clear that any trivial ball-arc pair is equivalent to any other, since the Reidemeister moves and isotopies of diagrams generate knot equivalence.
I used the Reidemeister moves as a tool to deal with $S^3$ isotopies, but this was not necessary.  Other options include (1) taking the ball-arc pair and extending it into a closed tubular neighborhood of $K$, then using the fact that every parameterization of a closed tubular neighborhood is isotopic to any other, and then finishing it off with the fact that any closed interval on $S^1$ is isotopic to any other, (2) using the parameterization of the trivial ball-arc pair to shrink it so that it lies inside a given tubular neighborhood, then seeing the way in which it sits with respect the foliation of the tubular neighborhood by disks to isotope it (depending on the present singularities) into a standard ball with only a maximum and minimum, and any pair of such balls are more-or-less-obviously isotopic.  Most of the complexity is keeping track of the arc inside the ball.
