Results 1 to 8 of 8

Math Help - exsistance of inverse?

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    128

    exsistance of inverse?

    hello!

    I am trying to work this out, not sure where to start even though the theorem appears intuitive! I thought I had proven it but then realised I has assumed the existance of an inverse!

    For functions f,g and sets A, B

    Given   f: A \rightarrow B and  g: B \rightarrow A and  g \circ f = 1_A, f \circ g = 1_B

    where 1_A is the function  1_A(a) = a,    \forall a \in A

    Then   f^{-1} exists and  g = f^{-1}

    To show the exsistance of an inverse requires that f is a bijection? not sure how to derive that from what is given.

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32
    Quote Originally Posted by matt.qmar View Post
    To show the existence of an inverse requires that f is a bijection?
    No. f need not have a 1-1 correspondence. take f(x) = c for some constant c \neq 0 The inverse would be g= 1/c
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by MacstersUndead View Post
    No. f need not have a 1-1 correspondence. take f(x) = c for some constant c \neq 0 The inverse would be g= 1/c
    No!

    If f(x) \equiv c, \ g(x) \equiv \frac{1}{c} for c \ne 0 then f(g(x)) = f \left( \frac{1}{c} \right) = c (but it should equal x!) and for the same reason g(f(x)) = \frac{1}{c}.

    @matt: By definition, a function has an inverse iff it is a bijection. You can try proving this here by contradiction (twice) - first suppose it is not injective and reach a contradiction, and then suppose it is not surjective and again reach a contradiction.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    MacstersUndead, when talking about functions, the "inverse" always refers to the inverse function, the function g(x) such that f(g(x))= x and g(f(x))= x, NOT the "reciprocal". A constant function does NOT have an inverse.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    128
    aha, ok! so, (not via contradiction though),
    for m,n \in A
    Assume  f(m) = f(n)
    then  g \circ  (f(m)) = g \circ (f(n))
    and since g \circ f = 1_A
    then  1_A(m) = 1_A(n)
    implies m=n
    since m,n \in A

    So f is 1-1

    Now to show it is onto,
    for b \in B
    f(g(b)) = 1_B(b) = b
    so for each b \in B
    there is a g(b) \in A
    so for each b \in B, \exists g(b) \in A
    such that f(g(b)) = b

    (each element of B is mapped onto)

    so f is onto,

    so we know f is a bijection
    so it has an inverse!

    now... how do we show the inverse is equal to g?

    thanks for the help everyone!!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    What properties does an inverse function have?
    What properties does g have?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Show that the inverse of f is the unique function such that f(f^{-1}(x))= x and f^{-1}(f(x))= x. Since you are told that g has those properities, it [b]is[/b the inverse.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2006
    Posts
    71
    Quote Originally Posted by matt.qmar View Post
    hello!

    I am trying to work this out, not sure where to start even though the theorem appears intuitive! I thought I had proven it but then realised I has assumed the existance of an inverse!

    For functions f,g and sets A, B

    Given   f: A \rightarrow B and  g: B \rightarrow A and  g \circ f = 1_A, f \circ g = 1_B

    where 1_A is the function  1_A(a) = a,    \forall a \in A

    Then   f^{-1} exists and  g = f^{-1}

    To show the exsistance of an inverse requires that f is a bijection? not sure how to derive that from what is given.

    Thanks in advance!
    Applying what you've been given to the definition of a one-sided inverse, you can conclude the following:

    1) From g \circ f = 1_A, g is acting as a left inverse of f and f is acting as a right inverse of g.
    2) From f \circ g = 1_B, f is acting as a left inverse of g and g is acting as a right inverse of f.

    Only a bijection, say f, has a unique two-sided inverse, i.e., a unique function that's both the left inverse and the right inverse of f.

    But from 1) and 2), you have that g is both a left and right inverse of f, and that f is both a left and right inverse of g.

    So on the one hand, you can say g = f^{-1}, the unique two-sided inverse of f.
    On the other hand, you can say f = g^{-1}, the unique two-sided inverse of g.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. left inverse equivalent to inverse
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 30th 2011, 03:58 PM
  2. Inverse tan and inverse of tanh
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 31st 2010, 06:20 AM
  3. secant inverse in terms of cosine inverse?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 30th 2010, 08:08 PM
  4. Replies: 1
    Last Post: April 9th 2010, 05:51 PM
  5. inverse trig values and finding inverse
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 6th 2009, 12:04 AM

Search Tags


/mathhelpforum @mathhelpforum