Suppose the relation R  A  B is a function

1. Inverse function. Suppose the relation R  A B is a function It means that for anya A there exists a uniqueb B, (a, b) R , or we have f (a)=b. Then the inverse relation R-1 = { (b, a) | bB, aA and f (a)=b}  BA also defines a function only if f : A B is a bijection.

2. R R-1   R is a function R-1is not a function. Why? R-1 R R is a function R-1is not a function. Why?

3. Theorem. Let f : A  B be a bijection. Then the inverse relation of f , defined from B to A as g = {(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.

4. f g (b) b B  A f a  g gf (a) The converse of the previous Theorem is also true. Theorem . Let f : 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 .

5. ? a1= a2 g(f (a1)= a1 a1 f (a1) = f (a2) g(f (a2)= a2 a2 Sketch of the proof that f is a bijection, given f : A  Band g : BA a A, g  f (a)= a b B, f  g (b)= b f is bijection = injection + surjection f is injection: f (a1) = f (a2)  a1= a2 Since g is a function,f (a1) = f (a2)  g(f (a1)=g(f (a2), i. e. a1= a2 .

6. Proof that if gf (a)= a for all aA , and fg (b)=b for all bB, then f is an injection. 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, g : BA. But gf (a1) = a1 and gf (a2)= a2 (gf (a)= a for all aA is given). It implies thata1 = a2 .

7. Sketch of the proof that f is a surjection, given f : A  Band g : BA a A, g  f (a)= a b B, f  g (b)= b 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

8. Proof that if gf (a)= a for all aA , and fg (b)=b for all bB, then f is surjection. 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.

9. ? 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

10. Proof that if gf (a)= a for all aA , and fg (b)=b for all bB, then g = f –1 . 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.

11. g  f (a2) a1 f (a1)= f (a2) g  f (a1) a2 b1 g f a • b2 Suppose f : A  Band g : BA and only one condition holds: a A, g  f (a)= a . Is it sufficient to imply any properties of g or f ? Can f be not injective? No. f must be injective. Can f be not surjective? Yes.

12. b1 f g a1 g •b2 a1 g b f a2• Suppose f : A  Band g : BA and only one condition holds: a A, g  f (a)= a . Is it sufficient to imply any properties of g or f ? Can g be not injective? Yes. No, g must be surjective. Can g be not surjective?

13. The following theorem shows how to compute the inverse of a composition. Theorem . If f : A  Band g : BC are two bijections, then (gf )-1 = f -1 g -1. Two functions f : A  Band g : BA are inverse of each other iff 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.

14. A B C g f g -1 f -1 gf Theorem . 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

15. 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

16. B f f (C) A CA Proofs involving functions. Let f: AB be 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 }.

17. 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.

18. 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}.

19. 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

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

21. 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.