# set theory-images and preimages (inverse images)

Prove that $f(f^{-1} (f(A) ) )=f(A)$.

I have proved the first part: Let $y\in f(f^{-1} (f(A) ) )$. Then there exists $x\in f^{-1} (f(A) )$ such that $f(x)=y$. By definition this means that $y=f(x)\in f(A)$.

How do I prove the other part?

• Can you not approach it the same way? – Clayton Aug 9 '13 at 18:28
• please can you help me in proving it. – sertar Aug 9 '13 at 18:34
• I can't shake the feeling that this question was asked sometime in the last two days. – Asaf Karagila Aug 9 '13 at 18:38

Let $x \in f(A)$. Then there is $y \in A$ such that $f(y)=x$. Since $y \in A$, $f(y) \in f(A)$ and so $y \in f^{-1}(f(A))$. Therefore, $f(y) \in f(f^{-1}(f(A)))$.