# Thread: Mapping

1. ## Mapping

Hey,

Given that $\displaystyle f: X \rightarrow Y$ and that $\displaystyle A\subset X$ and $\displaystyle B \subset Y$ then I have to show that:

$\displaystyle f(f^{-1}(B)) \varsubsetneq B$.

I have done the following:

Assume that $\displaystyle y \in f(f^{-1}(B)) \Leftrightarrow$
$\displaystyle x \in f^{-1}(B)$ such that $\displaystyle f(x)=y \Leftrightarrow$
$\displaystyle y=f(x) \in B \Leftrightarrow$
$\displaystyle f(f^{-1}(B)) \subset B$

What remains is to show that $\displaystyle f(f^{-1}(B)) \neq B$. Would a counterexample be sufficient?

Thanks.

2. ## Re: Mapping

Show that

if $\displaystyle B\subseteq f\left(X\right)$ then $\displaystyle f\left(f^{-1}(B)\right)=B$

3. ## Re: Mapping Originally Posted by detalosi Given that $\displaystyle f: X \rightarrow Y$ and that $\displaystyle A\subset X$ and $\displaystyle B \subset Y$ then I have to show that:
$\displaystyle f(f^{-1}(B)) \varsubsetneq B$.
Suppose $X=\{0,2,4,6,8\}~\&~Y=\{0,1,\cdots 9\}$ if $f(x)=x+1~\&~B=\{1,3,5\}$
$f\left(f^{-1}(B)\right)=?$

4. ## Re: Mapping

Yes I see that.
So I have to show that

$\displaystyle B \subseteq f(f^{-1}(B))$

does not hold.

I have followed the same "method" as before by assuming that $\displaystyle y \in B$ (and then concluding that $\displaystyle y \notin f(f^{-1}(B))$ but it does not lead towards what is desired!

How to move forward?

Plato, yes I understand the example. But I wish to move forward without a counterexample.

5. ## Re: Mapping Originally Posted by detalosi Yes I see that.
So I have to show that

$\displaystyle B \subseteq f(f^{-1}(B))$ does not hold.
NOTATION: In place of $f(f^{-1}(B))$ I use $\overrightarrow f (\overleftarrow f (B))$
Using the function I posted before, If $C=\{7,8,9\}$ then $\overleftarrow f (C)=\{6,8\}$
while $\overrightarrow f (\overleftarrow f (C))=\{7,9\}\ne C$
Are you sure you understand what exactly you are to show?

6. ## Re: Mapping

If $\displaystyle f(x)=x+1$ then $\displaystyle f^{-1}(x)=x-1$.

Then $\displaystyle f^{-1}(B) = \{ 0,2,4 \}$.

This means that $\displaystyle f(f^{-1}(B))=f(\{ 0,2,4 \})= \{ 1,3,5 \}$

Notation might be off.
Yes, I understand what I am trying to show. My thought process is as follows:

1) I want to show that $\displaystyle f(f^{-1}(B) \varsubsetneq B$

2) I have shown that $\displaystyle f(f^{-1}(B) \subseteq B$

With my question I simply intended to show that = does not hold and therefore the inclusion is strict??

7. ## Re: Mapping

Have I done enough to show what was intended?

8. ## Re: Mapping

you will not be able to show that

$$y\in B \Longrightarrow y\notin f\left(f^{-1}(B)\right)$$

simply because it is not true.

But the following is correct

if $y\in B$ and $y\notin f(X)$ then $y\notin f\left(f^{-1}(B)\right)$

and that is sufficient to prove that, in general, the inclusion is proper.

on the other hand, if $B\subseteq f(X)$ then you will not be able to find such a $y$ and $B=f\left(f^{-1}(B)\right)$