Since f maps from S to T (S-->T) and is invertible, then its inverse maps from T to S (T-->S).
Since the problem is to show that this inverse is unique, I guess my point was this:
f invertible means f is one to one. So if (s,t) is an ordered pair in the function f, then there is no OTHER ordered pair that has (s,t2), nor is there another ordered pair that has (s2, t). (This could be stated MUCH more formally, like I said... this is stream of consciousness... lol).
So (s,t) is the only ordered pair in f that involves those two specific elements of the domain S and the range T, right?
So the inverse, which swaps all the ordered pairs in f, contains (t,s) if f contains (s,t), right? And the uniqueness comes in the fact that, if (s,t) is in f, then (t,s) is in the inverse, and that holds for every unique point (s,t) in the original function f. So there is no "different" function that can be the inverse, because for EVERY point (s,t) in f the point (t,s) must be in the inverse... ??
Or something like that?? lol
sorry, like I said, it's been a while. I do love this kind of stuff, though. :)