Another prove/disprove; onto; inverse image

[dell500]

If A, B are sets and f: A -> B is onto, then f(f-1(C)) = C for any C subset B.

pf:
If x belongs to A, then f(x) belongs to B.
If B = f(A) for each y belongs to B and x belongs to A such that f(x) = y, then f(x) is a surjection (onto).
If f is a surjection, then every value in B has at least one pre-image in A.
That is, if y0 belongs to B, then there exists x0 belongs to A such that f(x0) = y0.
If C is a subset of B and f(x) belongs to B, then f(x) belongs to C.
So the inverse image, or pre-image, is f-1(C).
If f(x) belongs to C, then x belongs to C as well.
By definition of f o f-1(C) = f(f-1(C)).

I'm a bit stumped on where to go from here... Is my logic correct in saying all this as well?

Thanks

 

[pka]

If \(\displaystyle z \in C\) because of onto \(\displaystyle \left( {\exists p \in A} \right)\left[ {f(p) = z} \right]\).
But that means that \(\displaystyle p \in f^{ - 1} (C) \Rightarrow \;\left[ {z = f(p) \in f\left( {f^{ - 1} (C)} \right)} \right]\).
Now you work on the ‘other’ way.