Consider two well ordered sets P and Q.
A well ordered set is a set with a well ordering.
A well ordering is a linear ordering in which every subset of a set has a least element.
A linear ordering is a set ordered so that
for all p, q in P, either
p > q
q > p
or
p = q
It also has the property (as an ordering) that if
p < q and q < r then p = x for all x in W. (Prop 1)
Suppose this is not the case, thus the subset
X = {x in W such that f(x) < x}
is nonempty. If this is true then take the least element z. Suppose w = f(z) then w w and thus z is not the least element which is a contradiction.
Further, there exists only one automorphism on a set and the automorphism is the identity (Prop 2)
f and f^{-1} must be automorphic and thus order preserving, thus by (Prop 1) f(x) >= x and f^{-1}(x) >= x thus f(x) > x > f^{-1}(x) which is only true if f(x) = x = f^{-1}(x) which means that f is the identity function.
Now we prove (Claim 1):
Consider f(g^{-1}(x)) and g(f^{-1}(x)). These send W_2 to W_2 and thus must be the identity by (Prop 2). Thus let g^{-1}(f(x)) = h = f^{-1}(g(x)) and (g^{-1}(f(x))^{-1} = h^{-1} = f^{-1}(g(x)) = h, thus h = h^{-1}, and the only function with that property is of course the identity. Thus f^{-1}(g(x)) = I and g^{-1}(f(x)) = I, so f = g.
However, this makes much more sense intuitively.
Consider Prop 1. Suppose you had a train and you wanted to keep the train in the same order. In this example you could only move the train forward or backward or keep it where it is. But further consider the condition that every train must map to some other train. Thus either the caboose (minimal element of the whole set) must move up. It can’t move down any farther.
Consider Prop 2. Suppose you had a train, and you wanted to move every train car to another train car and preserve the order. Well, since the train cars are connected in a line (linear ordering) we can only move every train to itself.
Finally consider Claim 1. Suppose I have a set of colors. I have a function that takes the original train and it’s colors to the train colored with the new colors. Suppose I have two such functions. Then the second function must take the newly colorered train back to its original colors. Since it has to be the same colors as before in the same order, it is precisely the same, thus the composition of functions takes the train to itself, thus by prop 2 it is unique and the identity, thus those two functions are equal.
Cheers!
justamathguy
