Results 1 to 3 of 3

Math Help - injection & surjection proofs

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    7

    Question injection & surjection proofs

    Here is my problem...
    Let A, B, and C be nonempty sets and let f: AB and g: BC be functions.
    a) Prove that if g f is an injection, then f is an injection.
    b) If g f is an injection, then must g be an injection? Support your answer with a proof or counterexample.
    c) Prove that if g f is a surjection, then g is a surjection.
    d) If g f is a surjection, then must f be a surjection? Support your answer with a proof or a counterexpample.

    Any help is greatly appreciated! Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by ElenaS519 View Post
    Here is my problem...
    Let A, B, and C be nonempty sets and let f: AB and g: BC be functions.
    a) Prove that if g f is an injection, then f is an injection.
    b) If g f is an injection, then must g be an injection? Support your answer with a proof or counterexample.
    c) Prove that if g f is a surjection, then g is a surjection.
    d) If g f is a surjection, then must f be a surjection? Support your answer with a proof or a counterexpample.

    Any help is greatly appreciated! Thanks!

    Let : f(x)=f(y) ====> g(f(x))=g(f(x)) ====.> gof(x)=gof(y) =====> x=y

    since gof injective,hence f is injective

    let zεC THEN there exists xεA SUCH that gof(x) = z since gof surjective.

    BUT xεA THEN there exists yεB SUCH that f(x) =y ,hence gof(x)=g(f(x))=g(y)=z hence g surjective
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Let:

    f:{1,2}=A \Longrightarrow {a,b,c}=B, and

    g:{a,b.c}=B \Longrightarrow {d.e}=C

    AND such that:

    f(1) = a
    f(2) = b

    g(a) = d
    g(b) = e
    g(c} = e

    and now form:

    gof(1) = g(f(1)) =g(a)= d
    gof(2) = g(f(2)) =g(b) =e

    From the above we see that:

    gof is injection .surjection

    f is injection but g is not ,because g(b)=g(c) and b\neq c

    g is surjection but f is not ,because there is no element in A ,x such that f(x) =c
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Injection implied by Surjection
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: April 5th 2011, 08:16 PM
  2. Injection/Surjection
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 17th 2010, 10:13 PM
  3. Injection and Surjection
    Posted in the Advanced Algebra Forum
    Replies: 15
    Last Post: January 7th 2010, 06:44 AM
  4. injection, surjection and bijection questions
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 7th 2009, 10:23 AM
  5. Prove injection and surjection
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 31st 2009, 07:48 PM

Search Tags


/mathhelpforum @mathhelpforum