Today’s topics:

1. Today’s topics: • Properties of the composite functions. • Inverse functions. • Proofs with functions.

2. B g C f A a1 f (a1) g(f (a1))=g(f (a2)). a2 f (a2) Theorem 5. 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 Theorem 4, 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.

3. 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)

4. 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.

5. 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

6. x g f y Example. Let f : AB and g: B  C be two functions such that the composite function gf : AC is injective. Does it follow 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 injective  f injective

7. 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! 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.

8. g(y) = sin(y): [0, ]  [0, 1] gf =sin(x/2): [0, ]  [0, 1]

9. gf z x   y=f (x) What can be implied from surjective property of gf ? 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. Thus, for any zC there exists y B such that g(y)= z. Than means that g is surjective.

10. g f The following example shows that gf is surjective  f is surjective /

11. R-1 R 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? R is a function

12. R R-1   R-1is not a function. Why? R is a function Theorem 6. 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.

13. Proof. Since f : A  Bis a bijection, for eachb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.

14. B A f a   g b The converse of the Theorem 6 is also true. Theorem 7. Letf : A  Band g : BAbe two functions. If gf (a)= a for all aA , and fg (b)=b for all bB, then both f and g are bijections, and they are inverse functions of each other, that is f = g-1 and g = f –1 . We can first prove that if gf (a)= a for all aA , and fg (b)=b for all bB, then f is a bijection and g = f –1 .

15. ? a1= a2 a1 f (a1) = f (a2) a2 f is bijection = injection + surjection f is injection: f (a1) = f (a2)  a1= a2 g(f (a1)= a1 g(f (a2)= a2 Since g is a function,f (a1) = f (a2)  g(f (a1)=g(f (a2)

16. f is surjective: for any bB there exists aA such that f (a)=b a ? fg (b)=b g bB Since g is a function, a =g(b) is defined for bB f (a) = f (g (b))= fg (b)=b

17. ? g = f -1 definition of inverse f -1 (b)=a  f (a)=b ? g(b)=a  f (a)=b bB, g(b)=a  f (a)=b and a A, f (a)=b  g(b)=a Given:bB, fg (b)=b, so a=g(b), f (a)=b Given: aA, gf (a)=a, so b=f(a), g (b)=a

18. Proof. First prove that f is an injection, that is suppose f (a1)=f (a2) for a1, a2A. We want to show that a1 = a2 We have gf (a1) = gf (a2), since g is a function. But gf (a1) = a1 and gf (a2)= a2 (gf (a)= a for all aA is given). It implies thata1 = a2 . To prove that f is a surjection, take any bB to show that there always exists aA such that f (a)=b. Let a=g(b), then f (a)=f (g (b))=b, according to fg (b)=b for all bB. So, f is a bijection and f –1 defined as f –1 (b)=a  f (a) = b, is a function. To prove that g = f –1 , we need to prove that g(b)=a  f (a) = b. But fg (b)=b for any b B, that is g (b)=a  f (a)=b. In the same way,we havegf (a)= a for all aA, i. e. f (a)=b  g (b)=a . QED.

19. When two functions f : A  Band g : BA are inverse of each other, we have the relationship gf (a)=a for allaA and f g(b)=bfor allbB. It is equivalent to say, that gf =IA and f g(b)= IB, where IAdenotes identity function on A ( IA(a) = a for all aA ) and similarly, IB is identity function on B. The following theorem shows how to compute the inverse of a composition. Theorem 8. If f : A  Band g : BC are two bijections, then (gf )-1 = f -1 g -1. Proof. To prove that (gf )-1 = f -1 g -1, it suffices to prove that (f -1 g -1)(gf )=IA and (gf )(f -1 g -1) =IC

20. We have : (gf )(f -1 g -1) = ((gf )f -1 )g -1 by associative property of relation composition = (g(f f -1))g -1 ,by associative property = (gIB)g -1, since f f -1 =IB = gg -1, since gIB= g = IC Similarly we can prove that (f -1 g -1)(gf )=IA

21. B f f (C) A CA Proofs involving functions. Let f: AB is a function. Then for any subset of A, CA, we can define the set of images for elements of C that we denote as f (C): f (C) = {f (x) | xC }.

22. B f DB f-1(D) A For any subset of B, DB, we can define the set of pre-images of elements from D, which we denote as f-1(D): f-1(D) = {x | xA and f (x)D }. Pay attention that f-1(D) is just a notation for the set of pre-images, which does not assumes the existence of inverse function for f.

23. B A a x   b y   c z   Example: A = {a, b, c}, B ={x, y, z}, and f (a)=x, f (b)=y, f (c)=y. In this example, function f is neither injective nor surjective. So, inverse function f -1 does not exist (in other words, inverse relation of f is not a function). But we can find a set of pre-images for any subset of B. For example: f -1({z})=; f -1({x, z})={a}; f -1({x, y}) = {a, b, c}.

24. Questions • Is it always true, that f (A) = B? Ans. It is true only if f is surjective. Otherwise f (A)B. • Is it always true that f -1(B)=A? • Ans. Yes, since f is a function, any element of A has an image • that belongs B. As a result any element of A appears as a pre-image • of some element of B. • Is it true that if f(C)=f(D), where C, DA, then C=D? Ans. This is true only if f is injective. • Is it true that if f -1(C)=f -1(D), where C, D  B, then C=D? Ans. This is true only if f is surjective

25. Example. Prove that if f is injective, then XY= implies • f(X)f(Y)=. f B A X f (X) Y f (Y)

26. Proof by contradiction. Assume that XY= and f(X)f(Y), (1). (1) implies that there exists some common element, zf(X) and z f(Y), (2). From the definition of the set f(X) and f (Y) (2) implies that there exists xX, such that f(x)=z (3) and there exists yY, such that f(y)=z (4). So, we have f(x)= f(y)=z, (5) and since f is injective (5) implies that x=y, (6). Since xX and yY , (6) implies that XY in contradiction with assumption. The contradiction proves the initial statement.