I prefer the proof that takes f:n->n and gets you a labeled tree plus an ordered pair of vertices of the tree. It works as follows: If i is preperiodic for f, then connect i to f(i). Now put the periodic points in a line in increasing order, and permute them by f, so that the periodic point x with f(x) the lowest possible is first, etc. Then connect the periodic points up into a path in that order, and your ordered pair is the two ends of the path in order.
I prefer the proof that takes f:n->n and gets you a labeled tree plus an ordered pair of vertices of the tree. It works as follows: If i is preperiodic for f, then connect i to f(i). Now put the periodic points in a line in increasing order, and permute them by f, so that the periodic point x with f(x) the lowest possible is first, etc. Then connect the periodic points up into a path in that order, and your ordered pair is the two ends of the path in order.
Thank you for sharing that alternate method eyalminsky!
amazing simple sweet and intutive explanation thanks a lot!!😀