{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sole4 - ANALYSIS EXERCISE 4SOLUTIONS 1(a Let x1 x2 X be...

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

View Full Document Right Arrow Icon
ANALYSIS EXERCISE 4–SOLUTIONS 1. (a) Let x 1 , x 2 X be arbitrary. Suppose that g ( f ( x 1 )) = g ( f ( x 2 )). Then f ( x 1 ) = f ( x 2 ) since g is injective. Therefore x 1 = x 2 since f is injective. This proves that g f is injective. (b) Let z Z be arbitrary. Then y Y such that g ( y ) = z since g is surjective. Also x X such that f ( x ) = y since f is surjective. Therefore we have proved ( z Z )( x X )[ g ( f ( x )) = z ] . So g f is surjective. 2. (a) Let z Z . Then ( x X )[( g f )( x ) = z ] as g f is surjective. So g ( y ) = z where Y 3 y = f ( x ), thus g : Y Z is surjective. (a) Let y Y . Put z := g ( y ) Z . As g f is surjective, ( x X )[( g f )( x ) = z ]. But then z = g ( y ) = ( g f )( x ) = g ( f ( x )) . As g : Y Z is injective, it follows that y = f ( x ). In other words, ( x X )[ f ( x ) = y ] and hence f is surjective. 3. The composition g f : R - {- 1 , 1 } → R is defined by ( g f )( x ) = 1 f ( x ) = x - 1 x + 1 . 4. (a) f is injective since ( x 1 +1 x 1 - 1 = x 2 +1 x 2 - 1 ) ( x 1 = x 2 ). Let y X be arbitrary. We have to show that the equation x +1 x - 1 = y has a solution x X . Direct computation show that
Image of page 1

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

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

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern