Prove that T(a,b) is a permutation by directly computing T(a,b) ° T(a,b).

Let X be any set and fix a,b be an element of X. Define T(a,b):X to X by

T(a,b) = x if x is not an element of {a,b}
b if x = a
a if x = b

Prove that T(a,b) is a permutation by directly computing T(a,b) ° T(a,b).

Thanks a lot

You may be slightly confused by the notation here; T(a,b) is a function from X to X. The cases you write out are defining the value of the function T(a,b) at an element x in X. So it's not "T(a,b)" that equals x, or b, or a, but "the value of the function T(a,b) at x" which if you use the most common notation for function application would be written T(a,b)(x). But this may look confusing, as the parentheses surrounding (a,b) are part of the name of the function and only the parentheses surrounding x denote function application. So in answering I am going to denote the function T(a,b) by f. Namely, you have fixed elements a and b of x, and defined a function f from X to X by

f(x) = x if x is not in {a,b},
f(b) = a,
f(a) = b.

We are told to compute f o f. Well, if x is not an element of {a,b} we have

f(f(x)) = f(x) [by the definition of f: because x isn't in {a,b}, f(x) = x]
= x [by the definition of f, again, because x isn't in {a,b}, f(x) = x].

We also have

f(f(a)) = f(b) [by the definition of f: f(a) = b]
= a [by the definition of f, again: f(b) = a].

and

f(f(b)) = f(a) [by the definition of f: f(b) = a]
= b [by the definition of f, again: f(a) =b].

What this shows is that f(f(t)) = t for all t in X. In all cases--- it doesn't matter whether or not t is in {a,b}--- the calculation shows that when we apply f twice to t, we get t back.

This shows that f is one-to-one (if f(x) = f(y), applying f to both sides you deduce f(f(x)) = f(f(y)) and because f(f(x)) = x and f(f(y)) = y, you deduce that x = y) and onto (given x in X, we see that x is in the range of f because f(f(x)) = x). Slightly more glibly, f is one-to-one and onto because the fact that f(f(x)) = x for all x in X shows that f is invertible: in fact, f is its own inverse.

In any case, that's why f is a permutation of X.