Basic proof of onto functions

"Suppose $f: A \to B$ is onto. Show that $\forall \, Y \subset B, f(f^{-1}(Y)) = Y$.

Here's what I did:

1. Observe that $\forall x \in f^{-1}(Y), f(x) \in Y$. Then $f(f^{-1}(Y)) \subset Y$.

2. Since *f* is onto, $Y \subset f(A)$.

This is where I'm stuck. How do I continue step 2 to show that $f(f^{-1}(Y)) \supset Y$? Obviously $f^{-1}(Y) \subset A$, but I don't see how that helps...

Re: Basic proof of onto functions

Quote:

Originally Posted by

**phys251** "Suppose $f: A \to B$ is onto. Show that $\forall \, Y \subset B, f(f^{-1}(Y)) = Y$.

1. Observe that $\forall x \in f^{-1}(Y), f(x) \in Y$. Then $f(f^{-1}(Y)) \subset Y$.

2. Since *f* is onto, $Y \subset f(A)$.

Suppose $t\in Y$ then $\exists b\in A $ such that $f(b)=t$. Does that mean $t\in f(A)~?$

Re: Basic proof of onto functions

Let $ y \in Y $. Since $f$ is onto, there exists $ a \in A $ such that $\displaystyle f(a) = y $. Hence, $\displaystyle a\in f^{-1}(Y)$ implies $\displaystyle y = f(a) \in f(f^{-1}(Y))$. This shows $\displaystyle Y\subseteq f(f^{-1}(Y))$.