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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- June 3rd 2007, 04:09 PMObstacle1domain 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?

- June 3rd 2007, 04:54 PMThePerfectHacker
- June 4th 2007, 03:29 AMObstacle1
oh. I meant to write the domain and codomain of its inverse...

- June 4th 2007, 08:50 AMJhevon
yes, they are still the same

a function 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 - June 4th 2007, 09:26 AMObstacle1
- June 4th 2007, 09:42 AMPlato
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 there are two new functions and

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. - June 4th 2007, 09:53 AMJhevon
- June 4th 2007, 10:20 AMPlato
- June 4th 2007, 10:22 AMJhevon
- June 4th 2007, 10:27 AMPlato
- June 4th 2007, 11:06 AMObstacle1
- June 4th 2007, 11:09 AMJhevon
- June 4th 2007, 11:48 AMObstacle1
Here's what i originally read:

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

This set is given by = {a A: (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, (Y) = {a A: (a) Y}

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

: P(B)->P(A) - June 4th 2007, 12:52 PMPlato
What your text has given exactly what I posted above and used the symbol instead of . Many of us object to that last notation because it is easily confused confused with the reciprocal. Above you said that that text said was not no longer is wide use. Well I know that the text is mistaken with that judgment. Do you have any idea why 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 mappings from A to B. And that idea works well in transfinite arithmetic.

- June 4th 2007, 01:12 PMObstacle1