Results 1 to 5 of 5

Thread: identity function

  1. #1
    Banned
    Joined
    Sep 2009
    Posts
    502

    identity function

    I found the following from a book with no proof:

    Let f:A \rightarrow A be defined by the formula f(x)=x, then f is called the identity function, denoted by 1 or by 1_A.

    Let f:A \rightarrow B and it has the inverse function f^{-1}:B\rightarrow A, then f ^{-1}\circ f=1

    Question: Is it true that f ^{-1}\circ f=1?

    Isn't f(x)^{-1} \circ f(x)=x?
    Last edited by novice; May 20th 2010 at 12:25 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,492
    Thanks
    2730
    Awards
    1
    Quote Originally Posted by novice View Post
    I found the following from a book with no proof:

    Let f:A \rightarrow A be defined by the formula f(x)=x, then f is called the identity function, denoted by 1 or by 1_A.

    Let f:A \rightarrow B and it has the inverse function f^{-1}:B\rightarrow A, then f \circ f^{-1}=1

    Question: Is it true that f \circ f^{-1}=1?

    Isn't f(x) \circ f^{-1}(x)=x?
    Are you sure that you have not turned things around here.
    For one, if f:A\to B then f \circ f^{-1}:B\to B.
    So what are you asking?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Plato View Post
    Are you sure that you have not turned things around here.
    For one, if f:A\to B then f \circ f^{-1}:B\to B.
    So what are you asking?
    Yes, made a mistake. Sorry.

    Question: Is it true that f^{-1}\circ f =1?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,492
    Thanks
    2730
    Awards
    1
    But it is just a matter of notation: f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A .
    Recall that \left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right] thus f^{-1}\circ f(x)=1_A(x)=x.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Plato View Post
    But it is just a matter of notation: f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A .
    Recall that \left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right] thus f^{-1}\circ f(x)=1_A(x)=x.
    I found i_A a much better notation. Perhaps, the book is too old. The inverse function denoted by 1 is terribly confusing. I have mistaken it as a number.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hyperbolic function identity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 25th 2010, 09:26 PM
  2. an identity of floor function
    Posted in the Number Theory Forum
    Replies: 9
    Last Post: Aug 8th 2010, 09:36 AM
  3. arbitrary even function identity
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: Jul 28th 2009, 01:13 PM
  4. floor function - An Identity
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 3rd 2009, 08:28 AM
  5. hyperbolic function identity
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 24th 2007, 01:32 PM

Search Tags


/mathhelpforum @mathhelpforum