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. 
 A: *

*$\exists$ 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.

A: Note that:


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

*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 $y=f(x)\subset f(f^{-1}(y)) \subset f(f^{-1}(B)).$
A: Given $f : X \to Y$, and $\forall y \in Y \exists x \in X (f(x) = y)$.
Want to show that $\forall B \subset Y, f(f^{-1}(B)) = B$
We know that $f^{-1}(B) = \{ b : f(b) \in B \}$. If this set is never empty, then we have our result. The set is never empty by our second assumption.
A: First, $f^{-1}(y) = \lbrace x \in X : f(x) = y)\rbrace $. If $y \in B$, so exist $x \in X$ that $f(x) = y$ (f is onto), then $f(f^{-1}(y)) \in B$, by definition, so $f(f^{-1}(y)) \subset B$. Now, $y \in  f(f^{-1}(B))$, how f is onto exist $x \in f^{-1}(B)$, such that $f(x) = y$, then $y \in B$, so $f(f^{-1}(B)) \subset B$.
