Logic as a tool a guide to formal logical reasoning ( PDFDrive ) 365

Answers and Solutions to Selected Exercises 341 Exercises on functions 5.2.10 Since f is surjective, we have rng(f ) = B = dom(f −1 ) To prove that f −1 is injective, suppose that b1 , b2 ∈ B and that f −1 (b1 ) = f −1 (b2 ) Since rng(f ) = B , there are a1 , a2 ∈ A such that f (a1 ) = b1 and f (a2 ) = b2 Thus a1 = f −1 (f (a1 )) = f −1 (b1 ) = f −1 (b2 ) = f −1 (f (a2 )) = a2 Since we now have a1 = a2 , the injectivity of f allows us to conclude that f (a1 ) = f (a2 ) and therefore that b1 = b2 , as desired To show that f −1 is surjective, suppose that a ∈ A We must show that there is a b ∈ B such that f −1 (b) = a Take f (a) ∈ B : we have that f −1 (f (a)) = a 5.2.12 (a) Suppose f and g are injective, and that a1 , a2 ∈ A such that gf (a1 ) = gf (a2 ) We must show that a1 = a2 Thus g (f (a1 )) = g (f (a2 )) But since g is injective this means that f (a1 ) = f (a2 ), which in turn means that a1 = a2 , since f is injective (c) Follows immediately from Proposition 194, (a) and (b) (e) Suppose that gf is surjective and that c ∈ C We need to find an b ∈ B such that g (b) = c By the surjectivity of gf there is an a ∈ A such that gf (a) = c, that is, g (f (a)) = c Then f (a) ∈ B is the desired element of B 5.2.13 By Proposition 194(3) gf is bijective and therefore we know that it has an inverse (gf )−1 Clearly dom(f −1 g −1 ) = C = dom((gf )−1 ) and rng(f −1 g −1 ) = A = rng((gf )−1 ) We need to show that f −1 g−1 (c) = (gf )−1 (c) for all c ∈ C To this end, let c ∈ C arbitrarily Then, because gf is surjective, there is an a ∈ A such that gf (a) = c By definition of inverses (gf )−1 (c) = (gf )−1 (gf (a)) = a Now f −1 g −1 (c) = f −1 g −1 (gf (a)) = f −1 g−1 (g (f (a))) = f −1 (g −1 (g (f (a)))) By the associativity of function composition (Proposition 193) this last expression is equal to f −1 ((g −1 g )(f (a))) = f −1 (f (a)) = a 5.2.14 (a) For the sake of definiteness, suppose that f : A → B To prove the left-to-right implication suppose that f is injective and take any mappings g1 : C → A and g2 : C → A Further suppose that g1 f = g2 f We must show that g1 = g2 So let c ∈ C arbitrarily Then f g1 (c) = f g2 (c), that is, f (g1 (c)) = f (g2 (c)) Since f is injective, it follows that g1 (c) = g2 (c) To prove the converse we will proceed by contraposition, so suppose that f is not injective Then there are a1 , a2 ∈ A such that a1 = a2 but f (a1 ) = f (a2 ) We need to show that the left cancellation rule does not hold For this it is sufficient to find two mappings g1 and g2 such that f g1 = f g2 but g1 = g2 Let g1 : A → A and g2 : A → A be maps such that g1 is the identity map on A, that is, g1 (a) = a for all a ∈ A, and g2 (a) = a for all a ∈ A − {a1 , a2 } but g (a1 ) = a2 and g (a2 ) = a1 Then it is easy to check that f g1 = f g2 , but g1 = g2