ex2revsol

ex2revsol - MAT 300, Mathematical Structures Selected...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MAT 300, Mathematical Structures Spring 2010 Selected Solutions for Exam 2 Review We answered all questions from part (II) in class, and also parts (b), (f), and (g) from part (III). The remaining parts of (III) are answered here. III. (a) Suppose f : A B and g : B C , and assume that g f is one-to-one and onto. Prove that f is one-to-one and g is onto. Proof: Assume that f : A B and g : B C are functions and g f is one-to-one and onto. First we will prove that f is one-to-one. Let x 1 , x 2 A and assume that f ( x 1 ) = f ( x 2 ). Then, since g is a function, g ( f ( x 1 )) = g ( f ( x 2 )). But g f is one-to-one. This means that whenever g ( f ( x 1 )) = g ( f ( x 2 )), we must have x 1 = x 2 . Since x 1 = x 2 whenever f ( x 1 ) = f ( x 2 ), we conclude that f is one-to-one. Now, we wish to prove that g is onto. Let z C . Since g f is onto, there exists some x A for which g ( f ( x )) = z . Let y = f ( x ) B . Then z = g ( f ( x )) = g ( y
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/04/2010 for the course MAT 300 taught by Professor Thieme during the Spring '07 term at ASU.

Page1 / 3

ex2revsol - MAT 300, Mathematical Structures Selected...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online