Abstract Algebra Theory and Applications - Computer Science ...

14 CHAPTER 0 PRELIMINARIESgiven by the matrixB =( ) 3 0,0 0then an inverse map would have to be of the formT −1B(x, y) = (ax + by, cx + dy)and(x, y) = T ◦ T −1B(x, y) = (3ax + 3by, 0)for all x and y. Clearly this is impossible because y might not be 0.Example 14. Given the permutationπ =( 1 2) 32 3 1on S = {1, 2, 3}, it is easy to see that the permutation defined byπ −1 =( 1 2) 33 1 2is the inverse of π. In fact, any bijective mapping possesses an inverse, aswe will see in the next theorem.Theorem 0.4 A mapping is invertible if and only if it is both one-to-oneand onto.Proof. Suppose first that f : A → B is invertible with inverse g : B → A.Then g ◦ f = id A is the identity map; that is, g(f(a)) = a. If a 1 , a 2 ∈ Awith f(a 1 ) = f(a 2 ), then a 1 = g(f(a 1 )) = g(f(a 2 )) = a 2 . Consequently, f isone-to-one. Now suppose that b ∈ B. To show that f is onto, it is necessaryto find an a ∈ A such that f(a) = b, but f(g(b)) = b with g(b) ∈ A. Leta = g(b).Now assume the converse; that is, let f be bijective. Let b ∈ B. Since fis onto, there exists an a ∈ A such that f(a) = b. Because f is one-to-one,a must be unique. Define g by letting g(b) = a. We have now constructedthe inverse of f.□