Results 1 to 3 of 3

Math Help - Show that f o g is one-to-one

  1. #1
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128

    Show that f o g is one-to-one

    (a) Show that the composition of two one-to-one functions, f and g, is one-to-one

    (b) Express (f o g)^{-1}in terms of f^{-1} and g^{-1}.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear 450081592,

    1) Suppose a,b \in{Dom (f_og)} such that,

    f_o{g(a)}=p and f_o{g(b)}=p

    Then, f[g(a)]=f[g(b)]

    g(a)=g(b) since f is an one to one function.

    a=b since g is an one to one function.

    Therefore f_o{g(a)}=f_o{g(b)}\Rightarrow{a=b}

    Hence f_o{g} is an one to one function.

    2) Supose g(a) = c and f(c) = b,

    Therefore, f_og(a)=b

    Since f_og is one to one it is invertible,

    (f_og)^{-1}(b)=a-----------A

    Since f and g are one to one they are invertible.

    g^{-1}(c)=a and f^{-1}(b)=c

    Therefore g^{-1}[f^{-1}(b)]=a-------B

    From A and B,

    g^{-1}_of^{-1}(b)=(f_og)^{-1}(b) \forall b\in Dom{(f_og)}
    Last edited by Sudharaka; January 24th 2010 at 07:24 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by 450081592 View Post
    (a) Show that the composition of two one-to-one functions, f and g, is one-to-one

    (b) Express (f o g)^{-1}in terms of f^{-1} and g^{-1}.
    Quote Originally Posted by Sudharaka View Post
    Dear 450081592,

    1) Suppose a,b \in{Dom (f_og)} such that,

    f_o{g(a)}=p and f_o{g(b)}=p

    Then, f[g(a)]=f[g(b)]

    g(a)=g(b) since f is an one to one function.

    a=b since g is an one to one function.

    Therefore f_o{g(a)}=f_o{g(b)}\Rightarrow{a=b}

    Hence f_o{g} is an one to one function.

    2) Supose g(a) = c and f(c) = b,

    Therefore, f_og(a)=b

    Since f_og is one to one it is invertible,

    (f_og)^{-1}(b)=a-----------A

    Since f and g are one to one they are invertible.

    g^{-1}(c)=a and f^{-1}(b)=c

    Therefore g^{-1}[f^{-1}(b)]=a-------B

    From A and B,

    g^{-1}_of^{-1}(b)=(f_og)^{-1}(b) \forall b\in Dom{(f_og)}
    Jeez you two, haha. Try... \circ f\circ g

    Also, for the second one there is a particularly nice result if your functions are both mappings from a set X into itself. It follows that f,g\in S_X (the permutation group on X) from where it follows from basic group theory that \left(ab\right)^{-1}=b^{-1}a^{-1}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 18th 2011, 10:31 AM
  2. Using the MVT to show this
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 25th 2010, 12:19 PM
  3. Show the sum
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 21st 2010, 04:28 AM
  4. Show that ln5 < 1 + 1/2 + 1/3 + 1/4.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 6th 2009, 02:29 AM
  5. how to show show this proof using MAX
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 14th 2009, 01:05 PM

Search Tags


/mathhelpforum @mathhelpforum