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What is the notation [itex]id_{ }[/itex]?

  1. Jul 23, 2012 #1
    Preface: Can't understand notation from 1st year Real Analysis textbook that they pulled out of thin air without defining.

    1. The thing statement that they're proving
    Let A and B be sets. Moreover, let [itex]f : A \rightarrow B[/itex] be some injective map. Calculate [itex]f^{-1} \circ f[/itex].

    SOLUTION: To every [itex]y \in f(A)[/itex], the map [itex]f^{-1}[/itex] associates the corresponding [itex]x \in A [/itex] which satisfies [itex]f(x) = y[/itex]. In particular, it associates with [itex]f(x)[/itex] the element [itex]x[/itex] for all [itex]x \in A[/itex]. Hence:

    [itex]f^{-1} \circ f = id_A[/itex], [itex]f \circ f^{-1} = id_{f(A)}[/itex],

    where for every set [itex]C[/itex] the corresponding map [itex]id_C : C \rightarrow C [/itex] is defined by:

    [itex]id_C (x) := C[/itex]

    for all [itex]x \in C[/itex].


    2. My confusion
    What's [itex]id[/itex]?
     
  2. jcsd
  3. Jul 23, 2012 #2

    HallsofIvy

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    The identity map- the function that maps every member of a set to itself. If you are taking "mathematical analysis" surely you have seen inverse functions before? And you should know that the definition of "inverse function" is that f-1(f(x))= f(f-1(x))= x?

    But I doubt that your text says "idC(x):= C". Surely it says "idC(x)= x for all [itex]x\in C[/itex]". Check that again.
     
  4. Jul 23, 2012 #3
    1) It does say what you supposed that it did not say. Must be a typo.

    Page 31 of Beyer's Calculus & Analysis: A combined approach.

    2) You've solve my problem. Thanks :)
     
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