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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- Jun 3rd 2007, 03:09 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?

- Jun 3rd 2007, 03:54 PMThePerfectHacker
- Jun 4th 2007, 02:29 AMObstacle1
oh. I meant to write the domain and codomain of its inverse...

- Jun 4th 2007, 07:50 AMJhevon
yes, they are still the same

a function $\displaystyle f:A \mapsto B$ 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 - Jun 4th 2007, 08:26 AMObstacle1
- Jun 4th 2007, 08:42 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 $\displaystyle f:A \mapsto B$ there are two new functions $\displaystyle \overleftarrow f :P(B) \mapsto P(A)$ and $\displaystyle \overrightarrow f :P(A) \mapsto P(B)$

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. - Jun 4th 2007, 08:53 AMJhevon
- Jun 4th 2007, 09:20 AMPlato
- Jun 4th 2007, 09:22 AMJhevon
- Jun 4th 2007, 09:27 AMPlato
- Jun 4th 2007, 10:06 AMObstacle1
- Jun 4th 2007, 10:09 AMJhevon
- Jun 4th 2007, 10:48 AMObstacle1
Here's what i originally read:

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

This set is given by $\displaystyle \theta^{-1}$ = {a$\displaystyle \epsilon$A:$\displaystyle \theta$(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, $\displaystyle \theta^{-1}$(Y) = {a$\displaystyle \epsilon$A:$\displaystyle \theta$(a)$\displaystyle \epsilon$Y}

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

$\displaystyle \theta^{-1}$: P(B)->P(A) - Jun 4th 2007, 11:52 AMPlato
What your text has given exactly what I posted above and used the symbol $\displaystyle \overleftarrow f $ instead of $\displaystyle 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 $\displaystyle B^A$ was not no longer is wide use. Well I know that the text is mistaken with that judgment. Do you have any idea why $\displaystyle B^A$ 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 $\displaystyle m^n$ mappings from A to B. And that idea works well in transfinite arithmetic.

- Jun 4th 2007, 12:12 PMObstacle1