COT3100Hmk06Sol - COT 3100 Fall 2002 Homework #6 Solutions...

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Homework #6 Solutions 1) Let f : A B and g: B C denote two functions. If both f and g are surjective, prove that the composition g ° f: A C is a surjection as well. To prove g f is surjective, we need to prove that for all elements c C, there exists an element a such that g f (a) = c. Let c be an arbitrary element from the set C. Since g is surjective, we know there exists an element b such that g(b) = c. Similarly, for this element b, since f is surjective, we know there exists an element a such that f(a) = b. Substituting, we have g(f(a)) = c. Thus, we have shown that for an arbitrarily chosen element c, we can always find an element a such that g f (a) = c. 2) Let g: A A be a bijection. For n 2, define g n = g ° g ° ... ° g, where g is composed with itself n times. Prove that for n 2, that (g n ) -1 = (g -1 ) n . Use induction on n. Base case: n=1. LHS = (g 1 ) -1 = g -1 , the inverse of g. RHS = (g -1 ) 1 = g -1 , also the inverse of g. Inductive hypothesis: Assume for an arbitrary value of n=k that (g k ) -1 = (g -1 ) k . Inductive step: Under this assumption we must show that (g k+1 ) -1 = (g -1 ) k+1 . (g k+1 ) -1 = (g ° g k ) -1 = (g k ) -1 ° g -1 , by the rule for the inverse of a function composition. = (g -1 ) k ° g -1 , by the inductive hypothesis = (g -1 ) k+1 , by the definition of function composition. This proves the inductive step so we can conclude that (g
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COT3100Hmk06Sol - COT 3100 Fall 2002 Homework #6 Solutions...

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