Homework 7 Solutions

# Homework 7 Solutions - ⇒ Assume that f X → Y is...

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MATH 74: HOMEWORK 7 COMMENTS (1) Let x 1 , x 2 X , and suppose that g f ( x 1 ) = g f ( x 2 ), i.e., g ( f ( x 1 )) = g ( f ( x 2 )). Since g : Y Z is injective, it follows that f ( x 1 ) = f ( x 2 ). Since f : X Y is injective, it follows that x 1 = x 2 . Hence g f : X Z is injective. (2) ( ) Assume that f : X Y is surjective. We must show that for all y Y the set G f X × { y } is not empty. Let y Y . Since f is surjective there exists x X such that f ( x ) = y . Thus the point ( x, y ) X × Y is an element of G f . Since ( x, y ) X × { y } as well, we have ( x, y ) G f X × { y } . Thus the set G f X × { y } is not empty (we just exhibited an element of it!). ( ) Now suppose that for all y Y , the set G f X × { y } is not empty. Let y Y . Since G f X × { y } 6 = we know ( x, y ) G f X × { y } . In particular ( x, y ) G f , which means y = f ( x ). Hence x is a preimage of y . Book Problem # 18 Let z Z . Since g : Y Z is surjective there exists y Y such that g ( y ) = z . Since f : X Y is surjective there exists x X such that f ( x ) = y . Hence g ( f ( x )) = z , which shows that x is a preimage of z . Book Problem # 19
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Unformatted text preview: ( ⇒ ) Assume that f : X → Y is surjective. Deﬁne a function g : Y → X by sending y to any preimage x of y under f . Note that x exists because f is surjective. We claim that g is a right inverse of f , i.e., f ◦ g = I Y . We must check that f ◦ g ( y ) = I Y ( y ) for all y ∈ Y . Now g ( y ) = x , where x satisﬁes f ( x ) = y (by deﬁnition of g ). Hence f ◦ g ( y ) = f ( g ( y )) = f ( x ) = y = I Y ( y ) . ( ⇐ ) Assume there exists g : Y → X such that f ◦ g = I Y . We must show that for all y ∈ Y there exists x ∈ X such that f ( x ) = y . Let y ∈ Y , and set x = g ( y ). Then f ( x ) = f ( g ( y )) = f ◦ g ( y ) = I Y ( y ) = y, and hence x is a preimage for y . Date : March 13th, 2008. 1...
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