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Thread: identity function

  1. #1
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    identity function

    I found the following from a book with no proof:

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

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

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

    Isn't $\displaystyle f(x)^{-1} \circ f(x)=x$?
    Last edited by novice; May 20th 2010 at 11:25 AM.
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  2. #2
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    Quote Originally Posted by novice View Post
    I found the following from a book with no proof:

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

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

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

    Isn't $\displaystyle f(x) \circ f^{-1}(x)=x$?
    Are you sure that you have not turned things around here.
    For one, if $\displaystyle f:A\to B$ then $\displaystyle f \circ f^{-1}:B\to B$.
    So what are you asking?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Are you sure that you have not turned things around here.
    For one, if $\displaystyle f:A\to B$ then $\displaystyle f \circ f^{-1}:B\to B$.
    So what are you asking?
    Yes, made a mistake. Sorry.

    Question: Is it true that $\displaystyle f^{-1}\circ f =1$?
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  4. #4
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    But it is just a matter of notation: $\displaystyle f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A $.
    Recall that $\displaystyle \left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right]$ thus $\displaystyle f^{-1}\circ f(x)=1_A(x)=x$.
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  5. #5
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    Quote Originally Posted by Plato View Post
    But it is just a matter of notation: $\displaystyle f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A $.
    Recall that $\displaystyle \left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right]$ thus $\displaystyle f^{-1}\circ f(x)=1_A(x)=x$.
    I found $\displaystyle i_A$ 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.
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