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Let $f: X \to Y$ and $A \subseteq X, B \subseteq Y$. I want to check the following statements:

(a) $f[f^{-1}[f[A]]]=f[A]$

(b) $f^{-1}[f[f^{-1}[B]]]=f^{-1}[B]$

where $f[A]$ is the image of $A$ and $f^{-1}[B]$ is the preimage of $B$.

I would answer "yes" to both.

$f[A]$ gives us the image, $f^{-1}[f[A]]$ extracts the elements for which $f$ returns $f[A]$, so $f^{-1}[f[A]]=A$ and $f[f^{-1}[f[A]]]=f[A]$.

Similarly, $f^{-1}[B]$ gives us the elements for which $f$ returns $B$, so $f[f^{-1}[B]]=B$ and $f^{-1}[f[f^{-1}[B]]]=f^{-1}[B]$.

Is there anything more to that?

Zelazny
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    $f^{-1}[f[A]]=A$ need not be true. Consider a noninjective function with $f(x)=f(y)$ for different $x,y$. Now, if $A={x}$, your claim doesn't hold true.

    The statements are nonetheless true, but you have to work more on the proof.

     – martin.koeberl Feb 01 '17 at 17:01
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    You can have a look at http://math.stackexchange.com/questions/746123/prove-that-ff-1-fx-fx and http://math.stackexchange.com/questions/1061602/prove-or-disprove-f-1ff-1y-f-1y Looking at your proof and the previous comment, http://math.stackexchange.com/questions/368990/why-f-1fa-not-a and http://math.stackexchange.com/questions/78110/is-f-1fa-a-always-true might also be interesting for you. – Martin Sleziak Feb 01 '17 at 17:25

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