# domain and codomain of an inverse map

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• Jun 3rd 2007, 03:09 PM
Obstacle1
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?
• Jun 3rd 2007, 03:54 PM
ThePerfectHacker
Quote:

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.

Quote:

Why aren't they simply B and A?
They are.
• Jun 4th 2007, 02:29 AM
Obstacle1
oh. I meant to write the domain and codomain of its inverse...
• Jun 4th 2007, 07:50 AM
Jhevon
Quote:

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
• Jun 4th 2007, 08:26 AM
Obstacle1
Quote:

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...
• Jun 4th 2007, 08:42 AM
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.
• Jun 4th 2007, 08:53 AM
Jhevon
Quote:

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? :D i vaguely remember seeing something like this before
• Jun 4th 2007, 09:20 AM
Plato
Quote:

Originally Posted by Jhevon
could you post them any way? I vaguely remember seeing something like this before

Here is a pdf file.
• Jun 4th 2007, 09:22 AM
Jhevon
Quote:

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?
• Jun 4th 2007, 09:27 AM
Plato
Quote:

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.
• Jun 4th 2007, 10:06 AM
Obstacle1
Quote:

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...
• Jun 4th 2007, 10:09 AM
Jhevon
Quote:

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" :D
• Jun 4th 2007, 10:48 AM
Obstacle1

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)
• Jun 4th 2007, 11:52 AM
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.
• Jun 4th 2007, 12:12 PM
Obstacle1
Quote:

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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