answer clearly and in detail please The first 3 parts of Theorem 2 are repeated

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answer clearly and in detail please

The first 3 parts of Theorem 2 are repeated below for your convenience. Also, recall that Theorem 2a was already proven in the Lesson. Prove Theorem 2, parts b and c. (Hint: once you have proven part b, you can use parts a and b to prove part c.) Theorem 2: Let F: X rightarrow Y and G: Y rightarrow Z be functions. Then If F and G are both 1 - 1 then G F is 1 - 1. If F and G are both onto then G F is onto. If F and G are both 1 - 1 correspondences then G F is a 1 - 1 correspondence. Let X = {1, 2, 3, 4, 5}, Let three functions be defined as follows: F:X rightarrow X with F (1) = 3, F (2) = 2, F (3) = 2, F (4) = 2 ,F (5) = 5 G:X rightarrow X with G (1) = 1, G (2) = 3, G (3) = 4, G (4) = 5, G (5) = 2 H:X rightarrow X with H (1) = 2, H (2) = 4, H (3) = 1, H (4) = 3, H (5) = 5 It is easier to do the problem if you write the functions in pair form: F = {(1,3), (2,2), (3,2), (4,2), (5,5)} G = {(1 , 1), (2,3), (3,4), (4,5), (5,2)} H = {(1,2), (2,4), (3,1), (4,3), (5,5)} Find each of the following. Give your answers as sets of ordered pairs.

Explanation / Answer

3.)

(a)

if f and g are 1-1 then if f(x) = f(y) and g(x) = g(y) we have that x = y, thus assume then g(f(x) = g(f(y)) but f(x) and f(y) is contained in the domain of g and using that g is 1 - 1 we have that f(x) = f(y) thus using the f is 1-1 we have that x = y and

g o f is 1-1

(b)

Since f and g are onto then there exist a y for every x contained in the domain of f such that f(x) = y and there exist some b for every a contained in the domain of g such that g(a) = b thus, g(f(x)) = g(y) = b since y = a for some a contained in the domain of g thus, there exist some b for every x contained in the domain of f

(c)

a 1-1 correspondence is a function that is both 1-1 and onto thus by parts (a) and (b) we have that g o f is a 1-1 correspondence

4.)

f(g(1)) = f(1) = 3    (1,3)

f(g(2)) = f(3) = 2    (2,2)

f(g(3)) = f(4) = 2    (3,2)

f(g(4)) = f(5) = 5    (4,5)

f(g(5)) = f(2) = 2    (5,2)

h(f(1)) = h(3) = 1    (1,1)

h(f(2)) = h(2) = 4    (2,4)

h(f(3)) = h(2) = 4    (3,4)

h(f(4)) = h(2) = 4    (4,4)

h(f(5)) = h(5) = 5    (5,5)

g(h(1)) = g(2) = 3   (1,3)

g(h(2)) = g(4) = 5   (2,5)

g(h(3)) = g(1) = 1   (3,1)

g(h(4)) = g(3) = 4   (4,4)

g(h(5)) = g(5) = 2   (5,2)

f(g(h(1))) = f(g(2)) = f(3) = 2   (1,2)

f(g(h(2))) = f(g(4)) = f(5) = 5   (2,5)

f(g(h(3))) = f(g(1)) = f(1) = 3   (3,3)

f(g(h(4))) = f(g(3)) = f(4) = 2   (4,2)

f(g(h(5))) = f(g(5)) = f(2) = 2   (5,2)

when dealing with inverses, the order of the order pair switches f --> (x,y) and f^(-1) --> (y,x) thus

f^(-1) is not a function since 2 goes to 2 and 3 but to list the expression order pairs we have

(3,1), (2,2), (3,2), (2,4), (5,5)

g^(-1) --> (1,1), (3,2), (4,3), (5,4), (2,5)

h^(-1) --> (2,1), (4,2), (1,3), (3,4), (5,5)

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