Prove $F(F^{-1}(B)) = B$ for onto function

Suppose that $f:X \to Y$ is an onto function. Prove that for all subsets $B$ subset of $Y$, $f(f^{-1}(B)) = B$. I don't know how to do this if the function is not also one to one, which it is not. Any help proving this would be greatly appreciated.

• $\exists$ is the symbol for "there exists."
• $\wedge$ is the symbol for "and."
• $f:X\twoheadrightarrow Y$ is notation for "$f$ is a surjective (onto) function from $X$ to $Y$."
• $\mathcal{P}(Y)$ is notation for "powerset of $Y$" or "the set of all subsets of $Y$."

• "$\subseteq$" becomes "$=$" for onto functions.

Note that:

1. For any $y\in f(f^{-1}(B)$, there exists $x \in f^{-1}(B)$ satisfying $y=f(x)$. Clearly $y=f(x) \in B$.

2. For any $y\in B$, since $f(x)$ is an onto function, there is $x\in X$ satisfying that $f(x)=y$, i.e.,$x\in f^{-1}(y)$. So $y=f(x)\subset f(f^{-1}(y)) \subset f(f^{-1}(B)).$