Results 1 to 3 of 3

Math Help - Need help with second part of proof

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    New York
    Posts
    55

    Need help with second part of proof

    Suppose f:A→B is a given function. Then there is a function g:f(A)→A sich that
    g o f=IA & f o g= I f(A), if and only if f is a one to one function and g=f-1
    Proof(→):
    Suppose f is 1-1 and g=f-1 then for all a elements of A, if f(a)=b then g(b)=a.
    So g(f(a))=g(b)=a.
    Thus g o f=IA, Similarly, f(g(b))=f(a)=b for any element b of f(A), so f o g=If(A).

    Need help with (←)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    147

    Re: Need help with second part of proof

    We're given g \circ f = I_A and f \circ g = I_{f(B)}, and want to show that f is 1-1, and then that g = f^{-1}.

    What does a function being 1-1 mean? It means that if two images of the function are equal, then the originating domain elements must also be equal. (y = x^2 isn't 1-1, because a^2 = b^2 doesn't force a = b. y = x^3 is 1-1, since a^3 = b^3 does force a = b.)

    Set it up, and then there's an obvious way to proceed. Like this:

    Suppose x_1, x_2 \in A, and f(x_1) = f(x_2).
    Then... (fill in the reasoning)...
    Therefore x_1 = x_2.
    Therefore f is 1-1.

    For g = f^{-1}, again, go to the definiton, which is that f^{-1}(f(x)) = x \ \forall \ x \in A.

    Like this:

    For all x in the domain, g(f(x)) = ...(fill in the reasoning)... =x.
    Therefore g = f^{-1}.
    Last edited by johnsomeone; October 10th 2012 at 11:05 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Need help with second part of proof

    Quote Originally Posted by franios View Post
    Need help with (←)
    \displaystyle\begin{aligned}f(x_1)=f(x_2)& \Rightarrow g[f(x_1)]=g[f(x_2)]\\&\Rightarrow (g\circ f)(x_1)=(g\circ f)(x_2)\\&\Rightarrow I_A(x_1)=I_A(x_2)\\&\Rightarrow x_1=x_2\\&\Rightarrow f\mbox{ is injective}\end{aligned}

    Try the rest.

    P.S. Sorry, I didn't see the previous post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parameterised Curve Proof Part 1
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 4th 2011, 01:27 PM
  2. [SOLVED] A small part of a chain rule proof I don't get
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: January 11th 2011, 04:03 AM
  3. I need to verify part of the proof involve Green's function.
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: September 12th 2010, 04:58 PM
  4. Proof of k-multiset coefficients, part 1
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: June 19th 2010, 04:55 AM
  5. Proof for Part 2 of the fundamental theorm
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 30th 2009, 06:18 AM

Search Tags


/mathhelpforum @mathhelpforum