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sols12 - 12 OCTOBER 11TH INVERSES INJECTIVE SURJECTIVE...

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12. OCTOBER 11TH: INVERSES. INJECTIVE, SURJECTIVE FUNCTIONS. 29 12. October 11th: Inverses. Injective, surjective functions. 12.1 . Let f : A B be a function, and assume that f has a left inverse g and a right inverse h . Prove that g = h . (Hint: Use Proposition 11.14.) Deduce that if f has a left and a right inverse, then it has a two-sided inverse. Answer: Since g is a left inverse of f , then g f = 1 A . Since h is a right inverse of f , then f h = 1 B . (Cf. Definition 12.1.) From the first equality, and Proposition 11.15, h = 1 A h = ( g f ) h . By the second equality, and again Proposition 11.15, g = g 1 B = g ( f h ) . On the other hand, g ( f h ) = ( g f ) h by Proposition 11.14. Putting all together, g = g ( f h ) = ( g f ) h = h , showing that g = h , as required. This function g = h is both a left and a right inverse for f , so f has a two-sided inverse in this case (Definition 12.1). 12.2 . Let f : A B and g : B C be two functions. Prove that if g f has a left inverse, then f has a left inverse. Prove that if g f has a right inverse, then g has a right inverse. Answer: Assume g f has a left inverse h . Then h ( g f ) = 1 A (Definition 12.1). By associativity (Proposition 11.14),
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