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(a) g is not injective but g f is injective (b) f is not surjective but g f is surjective Solution The same example works for both Let A = f1g, B = f1;2g, C = f1g, and f A !B by f(1) = 1 and g B !C by g(1) = g(2) = 1 Then g f A !C is de ned by (g f)(1) = 1 This map is a bijection from A = f1gto C = f1g, so is injective and surjective. 0 1 23 4 5 6 7 8 9 5 ;. Title Microsoft Word 0_0 Author ptsankova Created Date 8/25/16 AM.

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