key4sp00 - COT3100-01, Spring 2000 S. Lang Solution Key to...

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COT3100-01, Spring 2000 S. Lang Solution Key to Assignment #4 (40 pts.) 3/26/2000 1. (10 pts.) Consider a set A = {1, 2}, a set B = { a , b , c }, and a set C = { x , y }. Define a relation R A × B as R = {(1, a ), (1, c ), (2, b )}, a relation S B × C as S = {( a , x ), ( b , y ), ( c , x )}, and a relation T C × A as T = {( x , 1), ( y , 2)}. Now answer each of the following questions with a short explanation. (a) Is R a function? If so, is it an injection? Relation R is not a function because 1 A is related to two elements a and c in B (that is, both (1, a ) and (1, c ) R .) (b) Is R –1 a function? If so, is it a surjection? R –1 = {( a , 1), ( b , 2), ( c , 1)} is a function from B to A (because for each element of B , there is exactly one related element of A ). The function R –1 : B A is a surjection because its range equals {1, 2} = A . (c) Is S a function? Is S –1 a function? Relation S is a function from B to C (because for each element of B , there is exactly one related element of C ). S –1 = {( x , a ), ( x , c ), ( y , b )} C × B is not a function because element x C is related to two element a and c of B . (d) Is the composed relation S –1 R –1 a function? If so, is it an injection? Is it a surjection? S –1 R –1 = {( x , 1), ( y , 2)} C × A is a function from C to A (because for each element of C , there is exactly one related element of A ). It is a surjection because its range equals {1, 2} = A . (e) Is the composed relation T R S a function? If so, is it an injection? T R S = ( T R ) S = {( x , a ), ( x , c ), ( y , b )} S = {( x , x ), ( y , y )} is a function from C to C . It is also an injection because each pair of distinct elements ( x and y ) are mapped to distinct images ( x and y , respectively). 2. (4 pts.) Let f : A B and g : B C denote two functions. If the function g o f : A C is a surjection, and g is an injection, then prove the function f is a surjection. To prove f : A B is a surjection, let b B ---- (1), we need to prove there exists (i.e., to find) a A such that f ( a ) = b ---- (2). Since b B by (1), g ( b ) C ---- (3) because g : B C by assumption. Since g o f : A C is a surjection by assumption, so (3) implies there exists a A such that g o f ( a ) = g
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key4sp00 - COT3100-01, Spring 2000 S. Lang Solution Key to...

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