# Adv Alegbra HW Solutions 43 - 2.16(i Let f X → Y be a...

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43 Solution. Let z Z . Since g is surjective, there is y Y with g ( y ) = z ; since f is surjective, there is x X with f ( x ) = y .I t follows that ( g f )( x ) = g ( f ( x )) = g ( y ) = z , and so g f is surjective. (iii) If both f and g are bijective, prove that g f is bijective. Solution. By the f rst two parts, g f is both injective and surjec- tive (iv) If g f is a bijection, prove that f is an injection and g is a sur- jection. Solution. If h = ( gf ) 1 , then ( hg ) f = 1 and g ( fh ) = 1. By Lemma 2.9, the f rst equation gives f an injection while the sec- ond equation gives g a surjection. 2.15 (i) If f : ( π/ 2 ,π/ 2 ) R is de f ned by a 7→ tan a , then f has an inverse function g ; indeed, g = arctan. Solution. By calculus, arctan ( tan a ) = a and tan ( arctan x ) = x . (ii) Show that each of arcsin x and arccos x is an inverse function (of sin x and cos x , respectively) as de f ned in this section. Solution. Each of the other inverse trig functions satis f es equa- tions analogous to sin ( arcsin x ) = x and arcsin ( sin x ) = x , which shows that they are inverse functions as de f
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Unformatted text preview: 2.16 (i) Let f : X → Y be a function, and let { S i : i ∈ I } be a family of subsets of X . Prove that f ³ [ i ∈ I S i ´ = [ i ∈ I f ( S i ). Solution. Absent. (ii) If S 1 and S 2 are subsets of a set X , and if f : X → Y is a function, prove that f ( S 1 ∩ S 2 ) ⊆ f ( S 1 ) ∩ f ( S 2 ) . Give an example in which f ( S 1 ∩ S 2 ) ±= f ( S 1 ) ∩ f ( S 2 ) . Solution. Absent. (iii) If S 1 and S 2 are subsets of a set X , and if f : X → Y is an injec-tion, prove that f ( S 1 ∩ S 2 ) = f ( S 1 ) ∩ f ( S 2 ) . Solution. Absent. 2.17 Let f : X → Y be a function. (i) If B i ⊆ Y is a family of subsets of Y , prove that f − 1 ³ [ i B i ´ = [ i f − 1 ( B i ) and f − 1 ³ \ i B i ´ = \ i f − 1 ( B i ). Solution. Absent....
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