Today’s topics:

1. Today’s topics: • Proof that a given function is injective or surjective. • Composite functions • Properties of the composite functions. • Inverse functions. • Proofs with functions.

2. Let A =R{1} and f : A A be defined by the formula: Prove that f is injective and surjective. 1) injective: Let xand y be arbitrary elements of A such that f (x) = f (y ), (1). We must show that (1) implies x = y by assumption since (x1)0 and (y1)0 by algebra

3. 2) surjective: for any y there exists x A , such that f (x)=y . Take arbitrary y A and find x A from f (x)=y

4. B C f g A a    f (a)=b g (b)=c gf Composition of functions (Note, that in the context of function composition, the order in which the functions appear is backward, i. e. the rightmost function is applied first) Theorem. Let f : AB and g: BC be two functions. The compositionof f and g as relations defines a function gf : A C, such that gf (a)= g(f(a)).

5. B C f g A a    f (a)=b g (b)=c gf Proof. We need to prove that composite relation gf is a function. For this we must prove two things: 1) the composition gf relates each element aA to some c C. 2) an element c C assigned to a by gf is unique, so we can denote it c= g(f (a)). 1) existence: Let b=f (a) B. Let c=g(b) C. So, by the definition of composition of relations, (a, c)gf . Thus, cC [(a, c)gf ]

6. b1 g c1 f a b2 c2 2) uniqueness: Suppose (a, c1)gf and (a, c2)gf . Then by the definition of composition, b1B, such that (a, b1)f and (b1, c1)g, and b2B, such that(a, b2)f and (b2, c2)g. Since f is a function, there may be only one b B, such that (a, b)f, so b1= b2. Since g is a function, (b, c1)g and (b, c2)g imply that c1= c2. Thus, (gf )(a)= c= g (b) =g (f (a)).

7. g f f g g g f f • • Properties of the composite function • How the injective (surjective) property of gf depends • on properties of f and g

8. B g C f A a1 f (a1) g(f (a1))=g(f (a2)). a2 f (a2) Theorem. Consider functions f : AB and g: B C, and the composition gf : AC. a) If both f and g are injective, then gf is injective. b) If both f and g are surjective, then gf is surjective. c) If both f and g are bijective, then gf is bijective. Proof. a) Let a1 and a2 be arbitrary elements of A such that (gf )(a1) = (gf )(a2). By the previous Theorem, it implies that g (f (a1)) = g (f (a2)). Since g is injective it implies that f (a1) = f (a2), and since f is injective by assumption, a1 = a2.

9. C B A g f c    b a gf b) Take an arbitrary element cC. Since g is surjective, there exists bB such that g(b)=c. Similarly, since f is surjective,there exists aA such that f (a)=b. Then (gf )(a) = g(f(a)) = g(b) = c. So, gf is surjective. g surjective: c, b (g(b)=c) f surjective: b, a (f (a)=b)

10. c) If both f and g are bijective, then both f and g are injective and surjective. By parts a) and b) gf is injective and surjective, so gf is bijective. So, injective (surjective) properties of two functions, f and g are sufficient condition for injective (surjective) property of the composite function. The question arises whether both conditions are necessary as well. Let f : AB and g: B C be two functions and the composition gf : AC is injective. What properties of of f and g can be implied? What properties of f and g can be implied if gf is surjective?

11. (gf )(x) x g x y f (gf )(y) y Example. Let f : AB and g: B  C be two functions such that the composite function gf : AC is injective. Prove or disprove that both f and g are injective. Injective property of gf : AC implies if (gf )(x)=(gf )(y), then x = y. Let’s prove by contradiction that f must be injective. For this assume that gf is injective, but f is not injective. Then we can find x, yA such that x y and f (x)=f (y). By taking composite function for these x and y we have that g(f (x)) = g(f (y)) while x y. But this results to contradiction with injective property of composite function. g( f (x))= g( f (y)) gf injective  f injective

12. g f What about gf injective  g injective? This implication can be disproved by the following counterexample: g is not injective, but gf is! If fwas surjective, then we would not be able to disprove that gf injective  g injective!

13. g(y) = sin(y): [0, ]  [0, 1] gf =sin(x/2): [0, ]  [0, 1] Example: f : [0, ]  [0, ], f (x)=x/2. g : [0, ]  [0, 1], g (y)=sin(y), not injective gf =sin(x/2): [0, ]  [0, 1], is injective, but g is not.

14. gf z x   Example. Let f : AB and g: B  C be two functions such that the composite function gf : AC is surjective. Prove or disprove that both f and g are surjective. Let’s prove that if gf is surjective then g is surjective. Surjective property of gf implies that for any zC there exists x A such that (gf )(x)=z. It means that g(f (x))=z. Since f is a function, there exists a unique element y B such that y=f (x). But g(f (x))= g(y)= z. y=f (x) Thus, for any zC there exists y B such that g(y)= z. Than means that g is surjective.

15. g f The following example shows that gf is surjective  f is surjective / If we knew that g were injective, we could prove that gf is surjective  f is surjective !

16. Example. Let f : RR, and g : RR denote two functions, where R is the set of real numbers. Suppose gf (x)=2x+1 for all x R ( i. e. the composite function is both surjective and injective). a) Prove that the function f is injection. Let f (x)= f (y), we need to show that x = y. Since g is a function on a set of real numbers, f (x)= f (y) implies g(f (x))= g (f (y)), i. e. 2x+1 = 2y+1 (by assumption gf (x)=2x+1). So, it yields x = y . QED. b) Prove the function g is a surjection. Take arbitrary z R, we need to show that y R such that g(y)=z. But for any z R there always exist x, y R, x = (z1)/2 and y=f (x), so that g(y) = g (f (x)) = 2(z1)/2+1=z

17. R-1 R R is a function Inverse function. Suppose the relation R  A B is a function. The inverse relation was defined as: R-1 = { (b, a) | (a, b) R }  BA Is R-1 a function? R-1is not a function. Why?

18. R R-1   R is a function R-1is not a function. Why? Theorem. Let f : A  B be a bijection. Then the inverse relation of f , defined from B to A as {(b, a)| bB, aA and f (a)=b}, is a function from B to A such that gf (a)=a for all a A and fg (b)=b for all b B. The function g is called the inverse function of f and is denoted f –1.

19. Proof. Since f : A  Bis a bijection, for each bB there exists one and only one element aA such that f (a) =b . Thus, the relation from B to A, relating bB to its pre-image aA under f , is a function. That is we defined a function g : BAsuch that for any bB, g(b) =a iff f (a)=b. Thus, gf (a)= g(f (a)) = a Similarly, fg (b)=f (g (b))=b.