Permutations and Inverses

Definition Let A be a set. If f : A  A is a 1-1 correspondence then f is called a permutation of A. Notation: S(A): the set of all permutations of A M(A): the set of all mapping from A to A

Ex. Let A = {0, 1}. There are 4 mappings from A to A. f 1 (0) = f 1 (1) = 0; f 2 (0) = f 2 (1) = 1 f 3 (0) = 0, f 3 (1) = 1; f 4 (0) = 1, f 4 (1) = 0 Then M(A) = S(A) =

Definition Let A be a set. A mapping I A : A  A defined by I A (x) = x,  x  A is called an identity mapping on A. Note: I A o f(x) = I A (f(x)) = f(x) and f o I A (x) = f(I A (x)) = f(x). So I A o f = f = f o I A I A is the identity element on M(A) w.r.t. the composition of mapping.

Definition Let f  M(A), where A is a (non-empty) set. If there is a mapping g  M(A) such that g o f = I A = f o g, then we say g is the inverse of f w.r.t. the composition. Denote g = f  1. Ex. Let f : R  R defined by f(x) = x/2. Then f  1 (x) = 2x.

Ex. f and g: Z  Z defined by f(n) = 2n and g(n) = n/2 if n is even; g(n) = 4 if n is odd. g o f(n) = g(f(n)) = g(2n) = 2n/2 = n, for all n  Z. Thus g is a left inverse for f. f o g(n) = f(g(n)) = f(n/2) = 2(n/2) = n if n is even; f o g(n) = f(g(n)) = f(4) = 8 if n is odd. Thus g is not a right inverse of f. Hence g is not an inverse of f.

Lemma Let A be a nonempty set, f : A  A. Then f is 1-1  f has a left inverse. Pf:

Lemma Let A be a nonempty set, f : A  A. Then f is onto  f has a right inverse. Pf:

Ex. f and g: Z  Z defined by f(n) = 2n and g(n) = n/2 if n is even; g(n) = 4 if n is odd.

Theorem Let A be a nonempty set, f : A  A. Then f is invertible.  f is a permutation on A. Pf: