Equivalence Relations and Functions October 15, 2013 Week 13-14 1 Equivalence Relation A relation on a set X is a subset of the Cartesian product X £X. Whenever (x; y) 2 R we write xRy, and say that x is related to y by R. For (x; y) 62 R, we write xRy6 . De¯nition 1. A relation R on a set X is said to be an equivalence relation if (a) xRx for all x 2 X (reflexive). (b) If xRy, then yRx (symmetric). (c) If xRy and yRz, then xRz (transitive). Let X be a set and R an equivalence relation on X. It is quite common to denote the equivalence relation R by » if there is only one equivalence relation to be considered. So xRy becomes x » y. For each x 2 X we de¯ne [x] = fy 2 X j y » xg; called an equivalence class of »; each element of [x] is called a represen- tative of the class [x]. The collection all equivalence classes of » is called the quotient set of X modulo », denoted X=». De¯nition 2. A partition of set X is a collection P = fA1;:::;Akg of Sk disjoint nonempty subsets of X such that X = i=1 Ai. Proposition 3. Let X be a set. If » is an equivalence relation on X, then the collection f[x]: x 2 Xg of equivalence classes of » is a partition of X. Conversely, if P = fA1;:::;Akg be a partition of X, then the relation Sk 2 R = i=1 Ai is an equivalence relation on X. 1 Example 1. For any positive integer m, the congruence relation modulo m is an equivalence relation on the set Z of integers. The quotient set of Z modulo m is Z=»= f[0]; [1];:::; [m ¡ 1]g: Example 2. Let » be a relation on R2, de¯ned by (s; t) » (u; v) if v ¡ t = 2(u ¡ s). Example 3. Let f : X ! Y be a surjective function. De¯ne a relation » on X by x1 » x2 if f(x1) = f(x2). Then » is an equivalence relation on X. Moreover, the function f~ : X=»¡! Y; [x] 7! f(x) p is a bijection. For instance, f : R2 ! R by f(x; y) = x2 + y2. 2 Function and Inverse Function De¯nition 4. Let X and Y be sets. A function from X to Y is a rule f that assigns each element x of X a single element y in Y . We write the rule as f : X ! Y , and say that x is sent to y; we also write x 7! y or y = f(x). Functions are also known as maps or mappings. De¯nition 5. Let X and Y be nonempty sets. A function f : X ! Y is said to be ² injective if distinct elements are sent to distinct elements; ² surjective if for every element y 2 Y , there is an element x 2 X such that f(x) = y; ² bijective if f is both injective and surjective. If so it is called a one- to-one correspondence. De¯nition 6. Let X : X ! Y and g : Y ! Z be functions. The composi- tion of f and g is a function g ± f : X ! Z de¯ned by (g ± f)(x) = g(f(x)) for all x 2 X: Proposition 7. Let f : X ! Y , g : Y ! Z, h : Z ! W be functions. Then the composition functions h ± (g ± f) and (h ± g) ± f are the same. 2 Proof. For each x 2 X, (h ± (g ± f))(x) = h((g ± f)(x)) = h(g(f(x))) = (h ± g)(f(x)) = ((h ± g) ± f)(x): De¯nition 8. A function f : X ! Y is said to be invertible if there is a function g : Y ! X such that (g ± f)(x) = x for all x 2 X; (f ± g)(y) = y for all y 2 Y: If so, the function g is called an inverse of f, written g = f ¡1. Proposition 9. Let f : X ! Y be a function. (a) If f is invertible, then its inverse is unique. (b) The function f is invertible if and only if f is a bijection. Proof. (a) Let g1; g2 be functions g : Y ! X such that (gi ± f)(x) = x for all x 2 X and (f ± gi)(y) = y for all y 2 Y , i = 1; 2. Since y = f(g2(y)) and g1(f(x) = x, then g1(y) = g1(f(g2(y))) = g2(y) for all y 2 Y: (b) Let g be the inverse of f. For x1; x2 2 X, if f(x1) = f(x2), then x1 = g(f(x1)) = g(f(x2)) = x2. So f is injective. For any y 2 Y , the element g(y) is sent to y, in fact, f(g(y)) = y by de¯nition of inverse. Example 4. (a) f : R ! R by f(x) = x2 is neither injective nor surjective. (b) f : R ! R by f(x) = x3 is bijective; its inverse g : R ! R is given by p g(x) = 3 x. x (c) f : R ! R+ by f(x) = e is bijective, where R+ = fx 2 R j x > 0g; and ¡1 ¡1 its inverse is f : R+ ! R is given by f (x) = ln x. 2 (d) f : R ! R¸0 by f(x) = x is surjective, where R¸0 = fx 2 R j x ¸ 0g. (e) f : R ! R by f(x) = x2 is bijective; its inverse f ¡1 : R ! R is 1 ¸0 ¸0 p + given by f ¡1(x) = x. 2 (f) f : R·0 ! R¸0 by f(x) = x bijective, where R·0 = fx 2 R j x · 0g; its ¡1 ¡1 p inverse f : R+ ! R is given by f (x) = ¡ x. 3 Example 5. (a) The function f : R ! R by f(θ) = sin θ is neither injective ¼ ¼ nor surjective. However, the function g :[¡ 2 ; 2 ] ! [¡1; 1] is bijective; ¡1 ¼ ¼ ¡1 its inverse g :[¡1; 1] ! [¡ 2 ; 2 ] is denoted by g (x) = arcsin x. (b) The function f : R ! R by f(θ) = cos θ is neither injective nor surjective. However, the function g : [0; ¼] ! [¡1; 1] is bijective; its inverse g¡1 : [¡1; 1] ! [0; ¼] is denoted by g¡1(x) = arccos x. ¼ 3¼ (c) The function f :[ 2 ; 2 ] ! [¡1; 1] by f(θ) = sin θ is bijective. Its inverse ¡1 ¼ 3¼ ¡1 ¼ f :[¡1; 1] ! [ 2 ; 2 ] is given by f (x) = 2 + arccos x. S1 ¼ ¼ (d) The function f : n=¡1(¡ 2 ; 2 ) ! R by f(θ) = tan θ is neither injective nor surjective. ¼ ¼ (e) The function f :(¡ 2 ; 2 ) ! R by f(θ) = tan θ is bijective; its inverse ¡1 ¼ ¼ ¡1 f : R ! (¡ 2 ; 2 ) is denoted by f (x) = arctan x. The function f1 : ¼ 3¼ ¡1 ¼ 3¼ ( 2 ; 2 ) ! R by f(θ) = tan θ is also bijective; its inverse f1 : R ! ( 2 ; 2 ) ¡1 is given by f1 (x) = ¼ + arctan x. Example 6. The complex exponential function f : C ! C is de¯ned by f(z) = ez, where if z is written as z = x + iy then ez = ex+iy := exeiy = ex(cos y + i sin y): The function f is injective but not surjective. Example 7. Hyperbolic functions (a) The hyperbolic sine function is de¯ned as ex ¡ e¡x sinh : R ! R; sinh(x) = 2 is bijective. Its inverse is the function p arsinh : R ! R; arsinh(x) = ln(x + x2 + 1): p Let y = 1(ex ¡e¡x). Then e2x ¡2yex ¡1 = 0. Thus ex = 1(2y § 4y2 + 4) = p 2 p 2 p y § y2 + 1. Since y2 + 1 > y and ex > 0, we must have ex = y + y2 + 1. Hence p x = ln(y + y2 + 1): (b) The hyperbolic cosine function ex + e¡x cosh : R ! R; cosh(x) = 2 4 is neither injective nor surjective. It has no inverse function! However, the restriction f : R¸0 ! [1; 1); f(x) = cosh(x) is bijective and its inverse function is given by p f ¡1 : [1; 1) ! R; f ¡1(x) = arcosh(x) = ln(x + x2 ¡ 1): p Let y = 1(ex + e¡x). Then e2x ¡ 2yex + 1 = 0. Thus ex = 1(2y § 4y2 ¡ 4) = p 2 2 p p 2 x y § y ¡ 1.p Since e goes top1 when y goes to 1 and y + 1 > y ¡ 1, 2 then 0 < y ¡ py ¡ 1 = y ¡ (y + 1)(y ¡ 1) < y ¡ (y ¡ 1) = 1, we must have ex = y + y2 ¡ 1. Hence p x = ln(y + y2 ¡ 1): The function g : R·0 ! [1; 1) by g(x) = cosh(x) is also bijective. Its inverse function is given by p ¡1 ¡1 2 g : [1; 1) ! R·0; g (x) = ¡ ln(x + x ¡ 1): (c) The hyperbolic tangent function sinh x ex ¡ e¡x e2x ¡ 1 tanh : R ! (¡1; 1); tanh = = = cosh x ex + e¡x e2x + 1 is bijective. Its inverse function is 1 1 + x artanh : (¡1; 1) ! R; artanh(x) = ln : 2 1 ¡ x e2x¡1 2x 2x 1+y 1 1+y Let y = e2x+1. Then (1 ¡ y)e = 1 + y, i.e., e = 1¡y . Thus x = 2 ln 1¡y . (d) The hyperbolic cotangent function coth : (¡1; 0) [ (0; 1) ! (¡1; ¡1) [ (1; 1); cosh x ex + e¡x e2x + 1 coth(x) = = = sinh x ex ¡ e¡x e2x ¡ 1 is bijective. Its inverse function is 1 x + 1 artanh : fx 2 R : jxj > 1g ! fx 2 R : jxj > 0g; arcoth(x) = ln : 2 x ¡ 1 Proposition 10. Let f : X ! Y be a function from a ¯nite set X to a ¯nite set Y .