# Math Help - domain and codomain of an inverse map

1. ## 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?

2. Originally Posted by Obstacle1
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.

3. oh. I meant to write the domain and codomain of its inverse...

4. Originally Posted by Obstacle1
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

5. Originally Posted by Jhevon
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...

6. 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.

7. Originally Posted by Plato
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

8. Originally Posted by Jhevon
could you post them any way? I vaguely remember seeing something like this before
Here is a pdf file.

9. Originally Posted by Plato
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?

10. Originally Posted by Jhevon
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.

11. Originally Posted by Jhevon
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...

12. Originally Posted by Obstacle1
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"

13. Here's what i originally read:

Given a map $\theta$:A->B, we can associate with any element b $\epsilon$B 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 $\epsilon$A: $\theta$(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 $\epsilon$A: $\theta$(a) $\epsilon$Y}

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

$\theta^{-1}$: P(B)->P(A)

14. 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.

15. Originally Posted by Plato
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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