# Thread: domain and codomain of an inverse map

1. 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 $\displaystyle f:A \mapsto B$ 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.
Originally Posted by Jhevon
do these functions have a name? are they called "power functions" or something like that?
Originally Posted by Plato
I do not know. I first saw these is a text by Charles C Pinter.
Originally Posted by Obstacle1

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)
In the (old) texts that I have,

$\displaystyle f(A) = \{f(x) \in Y | x \in A \subset X \}$

is called the image of the set $\displaystyle A$ and

$\displaystyle f^{-1} (B) = \{ x \in X | f(x) \in B \subset Y \}$

is called the pre-image of $\displaystyle B$ and all the facts in Plato's PDF document are proven with these definitions. Pinter turns these into functions on the power sets. I wonder why? What use is made of image and pre-image as functions? Do we get anything extra out of this idea? (I guess this could be just a teaching device to drive home the fact that the inputs and outputs to the image and pre-image are sets. Or maybe it's a better notation.)

2. Originally Posted by JakeD
In the (old) texts that I have,

$\displaystyle f(A) = \{f(x) \in Y | x \in A \subset X \}$

is called the image of the set $\displaystyle A$ and

$\displaystyle f^{-1} (B) = \{ x \in X | f(x) \in B \subset Y \}$
Why do you get the impression that it is old.

My book on Abstract Algebra uses this notation as well, except it uses $\displaystyle f[A]$ instead of $\displaystyle f(A)$.

3. Originally Posted by JakeD
In the (old) texts that I have,

$\displaystyle f(A) = \{f(x) \in Y | x \in A \subset X \}$

is called the image of the set $\displaystyle A$ and

$\displaystyle f^{-1} (B) = \{ x \in X | f(x) \in B \subset Y \}$

is called the pre-image of $\displaystyle B$ and all the facts in Plato's PDF document are proven with these definitions. Pinter turns these into functions on the power sets. I wonder why? What use is made of image and pre-image as functions? Do we get anything extra out of this idea? (I guess this could be just a teaching device to drive home the fact that the inputs and outputs to the image and pre-image are sets. Or maybe it's a better notation.)
Originally Posted by ThePerfectHacker
Why do you get the impression that it is old.

My book on Abstract Algebra uses this notation as well, except it uses $\displaystyle f[A]$ instead of $\displaystyle f(A)$.
I was just pointing out my texts aren't current, you know, like old.

So I went back to my old texts and found the most precise one, the reference text, does define the image and pre-image as maps between the power sets. It calls these "the induced maps." But it doesn't introduce any new notation or prove anything different than the others. So I conclude this is the most precise definition, but nothing very substantial comes out of it.

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