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Archangel Macsika

Suppose g : A → B and f : B → C are functions. a. Show that if f ◦g is onto, then f must also be onto. b. Show that if f ◦g is one-to-one, then g must also be one-to-one. c. Show that if f ◦g is a bijection, then g is onto if and only if f is one-to-one.

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Suppose g : A\to B and f : B\to C  are functions.

a. Show that if f\circ g  is onto, then f  must also be onto.

Answer. Assume that f\circ g  is onto. Let c\in C . Since f\circ g  is onto, there is a\in A  such that c = (f\circ g)(a)  by the definition of onto function. By the definition of function composition, c = f(g(a)) . Let b = g(a) . This b  belongs to B  and is such that c = f(b). Therefore, for every c\in C , there is b\in B  such that c = f(b) . By the definition of onto function, f  is onto.

b. Show that if f\circ g  is one-to-one, then g  must also be one-to-one.

Answer. Let a_1, a_2\in A  such that g(a_1) = g(a_2) . Then

(f\circ g)(a_1) = f(g(a_1)) = f(g(a_2))= (f\circ g)(a_2).

 Since f\circ g  is one-to-one, a_1=a_2 . Therefore, for all a_1, a_2\in A  such that g(a_1) = g(a_2) , a_1=a_2 . By the definition of one-to-one function, g  is one-to-one.

c. Show that if f\circ g  is a bijection, then g  is onto if and only if f  is one-to-one.

Answer. Assume that f\circ g  is a bijection.

Assume that g  is onto. Let b_1, b_2\in B  such that f(b_1) = f(b_2) . Since g  is onto, by the definition of onto function, there are a_1, a_2\in A  such that b_1 = g(a_1)  and b_2 = g(a_2) . We have

(f\circ g)(a_1) = f(g(a_1)) = f(b_1)

= f(b_2) = f(g(a_2)) = (f\circ g)(a_2).

 Since f\circ g  is a bijection, it is one-to-one, hence a_1=a_2 , and

b_1 = g(a_1) = g(a_2) = b_2.

  •  Therefore, for all b_1, b_2\in B  such that f(b_1) = f(b_2) , b_1=b_2 . By the definition of one-to-one function, f  is one-to-one.
  • Assume that f  is one-to-one. Let b\in B . Since f\circ g  is a bijection, it is onto, hence there is a\in A  such that f(b) = (f\circ g)(a) = f(g(a)) . Since f  is one-to-one, b = g(a) . Therefore, for every b\in B , there is a\in A  such that b = g(a) . By the definition of onto function, g  is onto.

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