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Math Help - domain and codomain of an inverse map

  1. #1
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    domain and codomain of an inverse map

    If you have a map f:A->B , i've read that its domain and codomain are the powersets of B and A respectively. Why is this so? Why aren't they simply B and A?
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    Quote Originally Posted by Obstacle1 View Post
    If you have a map f:A->B , i've read that its domain and codomain are the powersets of B and A respectively. Why is this so?
    I do not know what you are talking about.

    Why aren't they simply B and A?
    They are.
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    oh. I meant to write the domain and codomain of its inverse...
    Last edited by ThePerfectHacker; June 4th 2007 at 06:31 AM.
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    Quote Originally Posted by Obstacle1 View Post
    oh. I meant to write the domain and codomain of its inverse...
    yes, they are still the same

    a function f:A \mapsto B has domain A and codomain B, since the domain is the set of inputs and the codomain is the set of outputs. Now an inverse function is a function that does pretty much the reverse of f. as in it takes the output from f and brings it back to the original input. when you consider that, you will realize that it takes the output from f as its input and the input from f as its output, and therefore its domain would be B and it's codomain A. note that an invertible function will be bijective
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    Quote Originally Posted by Jhevon View Post
    yes, they are still the same

    a function f:A \mapsto B has domain A and codomain B, since the domain is the set of inputs and the codomain is the set of outputs. Now an inverse function is a function that does pretty much the reverse of f. as in it takes the output from f and brings it back to the original input. when you consider that, you will realize that it takes the output from f as its input and the input from f as its output, and therefore its domain would be B and it's codomain A. note that an invertible function will be bijective
    That's what i had previously thought, and it stands to reason. I'll review over what I read though as I presume i must have taken it out of context...
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    What Jhevon gave you is the standard and usual definition.
    However, some set theory texts have introduced two new types of functions.
    Given a function f:A \mapsto B there are two new functions \overleftarrow f :P(B) \mapsto P(A) and \overrightarrow f :P(A) \mapsto P(B)
    Could this be what you have in mind?
    The definitions are rather difficult to type even in TeX.
    If this is it, then I will post a pdf file containing the definitions and examples.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    What Jhevon gave you is the standard and usual definition.
    However, some set theory texts have introduced two new types of functions.
    Given a function f:A \mapsto B there are two new functions \overleftarrow f :P(B) \mapsto P(A) and \overrightarrow f :P(A) \mapsto P(B)
    Could this be what you have in mind?
    The definitions are rather difficult to type even in TeX.
    If this is it, then I will post a pdf file containing the definitions and examples.
    could you post them any way? i vaguely remember seeing something like this before
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    could you post them any way? I vaguely remember seeing something like this before
    Here is a pdf file.
    Attached Files Attached Files
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    What Jhevon gave you is the standard and usual definition.
    However, some set theory texts have introduced two new types of functions.
    Given a function f:A \mapsto B there are two new functions \overleftarrow f :P(B) \mapsto P(A) and \overrightarrow f :P(A) \mapsto P(B)
    Could this be what you have in mind?
    The definitions are rather difficult to type even in TeX.
    If this is it, then I will post a pdf file containing the definitions and examples.
    do these functions have a name? are they called "power functions" or something like that?
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    do these functions have a name? are they called "power functions" or something like that?
    I do not know. I first saw these is a text by Charles C Pinter.
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    Quote Originally Posted by Jhevon View Post
    do these functions have a name? are they called "power functions" or something like that?
    I thought power functions were simply functions of the form x^n with only one term (also called monomial).

    I'll write up what my textbook has in just a minute...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Obstacle1 View Post
    I thought power functions were simply functions of the form x^n with only one term (also called monomial).
    yes, you are correct. i was just blurting out words, i thought "power sets" --> "power functions"
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  13. #13
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    Here's what i originally read:

    Given a map \theta:A->B, we can associate with any element b \epsilonB the set of those elements of A that take the value b under \theta. This set is the pre-image of b under \theta and is written \theta^{-1}

    This set is given by \theta^{-1} = {a \epsilonA: \theta(a)=b}.

    The notation suggests that the inverse is a map with domain B but unless we are very lucky the codomain is not A. The above equation shows that the values of the inverse map are actually subsets of A so that the codomain is the powerset P(A).


    More generally, we can make the following definition:

    For all subsets Y of B, \theta^{-1}(Y) = {a \epsilonA: \theta(a) \epsilonY}

    of the pre-image of any subset Y contained in B. This delivers a map

    \theta^{-1}: P(B)->P(A)
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  14. #14
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    What your text has given exactly what I posted above and used the symbol \overleftarrow f instead of f^{ - 1} . Many of us object to that last notation because it is easily confused confused with the reciprocal. Above you said that that text said B^A was not no longer is wide use. Well I know that the text is mistaken with that judgment. Do you have any idea why B^A was chosen (probably by Cantor) for the set of the set of all mappings from A to B? Well, if A has n elements and B has m elements then there are m^n mappings from A to B. And that idea works well in transfinite arithmetic.
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  15. #15
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    Quote Originally Posted by Plato View Post
    What your text has given exactly what I posted above and used the symbol \overleftarrow f instead of f^{ - 1} . Many of us object to that last notation because it is easily confused confused with the reciprocal. Above you said that that text said B^A was not no longer is wide use. Well I know that the text is mistaken with that judgment. Do you have any idea why B^A was chosen (probably by Cantor) for the set of the set of all mappings from A to B? Well, if A has n elements and B has m elements then there are m^n mappings from A to B. And that idea works well in transfinite arithmetic.
    Yea i forgot to mention that it did justify that notation by showing that if |A|=a and |B|=b then |Map(A,B)|= b^a. (obviously the same as what you've written). I hadn't seen the \overleftarrow f notation before.
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