hw1sol - Math 113 Homework 1 Solutions 1. (a) Let f : X Y...

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Unformatted text preview: Math 113 Homework 1 Solutions 1. (a) Let f : X Y and g : Y Z . Show that if g f : X Z is onto, then g is onto. Assume g f is onto, and let z be an element of Z . Since g f is onto, there exists x X such that g f ( x ) = z . Let y = f ( x ) Y : then g ( y ) = g ( f ( x )) = z . Therefore, for every z Z there exists y Y such that g ( y ) = z . Hence g is onto. Show that if g f is one-to-one, then f is one-to-one. Assume g f is one-to-one, and let x 1 , x 2 X be such that f ( x 1 ) = f ( x 2 ). Then g f ( x 1 ) = g ( f ( x 1 )) = g ( f ( x 2 )) = g f ( x 2 ), and since g f is one-to-one we deduce that x 1 = x 2 . Hence f is one-to-one. (b) Show that f : X Y is bijective (i.e., one-to-one and onto) if and only if there exists g : Y X with g f = id X and f g = id Y . First we show that, if f is bijective, then there exists g with the stated properties. Given a bijective map f : X Y , define g : Y X as follows. Given y Y , there exists a unique x X such that f ( x ) = y (since f is bijective), and we define g ( y ) = x . Given x X , let y = f ( x ), then by definition g ( y ) = x so g ( f ( x )) = x . Hence g f = id X . Given y Y , let x = g ( y ), by definition f ( x ) = y , so f ( g ( y )) = y . Therefore f g = id Y . Conversely, assume there exists g : Y X such that f g = id Y and g f = id X . Then g f = id X is one-to-one, so by part (a) f is one-to-one. Moreover f g = id Y is onto, so by part (a) f is onto. Hence f is a bijection. (Optional: what happens if you drop one of the two conditions on g ?) If we can find g : Y X such that g f = id X , then by part (a) f is one-to-one. However f is not necessarily onto: for example consider X = { 1 } , Y = { 1 , 2 } , f (1) = 1, g (1) = g (2) = 1. Similarly, if we can find g such that f g = id Y then by (a) f is onto, but it is not necessarily one-to-one (same counterexample, exchanging X with Y and f with g ). 2. Fraleigh, section 0, exercises 2934. Determine whether the given relations are equivalence relations; describe the partition arising from each equivalence relation....
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This note was uploaded on 03/14/2012 for the course MATH 113 taught by Professor Ogus during the Spring '08 term at University of California, Berkeley.

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hw1sol - Math 113 Homework 1 Solutions 1. (a) Let f : X Y...

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