analysis - Prove: f is surjective -->$f(f^{-1}(S)) = S$

I have to prove this exercise for my math-study:

Let $f: X \rightarrow Y$ be a function and $S \subseteq Y$

Prove: $f$ is surjective $\Rightarrow$ $f(f^{-1}(S)) = S$

I divided this exercise in two parts,
first proving that $S \subseteq f(f^{-1}(S))$.
This is what I did:

Assume $f$ is surjective $\Rightarrow$ $\forall s$ $\in S$ $\exists x \in$ $f^{-1}(S)$ such that $f(x) = s \Rightarrow s$ $\in$ $f(f^{-1}(S))$ $\Rightarrow$

$$S \subseteq f(f^{-1}(S))$$

Is this part right, or did I make any mistakes?

For the second part, I have to prove that $f(f^{-1}(S)) \subseteq S$

I began with this:

Assume $x$ $\in$ $f(f^{-1}(S))$. $f^{-1}(S)$ = {$x$ $\in$ X | $f(x)$ $\in$ S}

But I don't know how to prove from that that $x \in S$. Could you please help me with these two questions? Thanks in advance!

Answer

If $x\in f(f^{-1}(S))$, then $x = f(y)$ for some $y\in f^{-1}(S)$. So $f(y)\in S$, i.e., $x\in S$.

Conversely, if $x\in S$, then since $f$ is surjective, there exists a $u\in X$ such that $f(u) = x$. So $f(u)\in S$, which implies $u\in f^{-1}(S)$. Therefore $x = f(u)\in f(f^{-1}(S))$.