# Basic proof of onto functions

• May 15th 2014, 12:49 PM
phys251
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...
• May 15th 2014, 01:11 PM
Plato
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 $t\in f(A)~?$
• May 15th 2014, 01:13 PM
SlipEternal
Re: Basic proof of onto functions
Let $y \in Y$. Since $f$ is onto, there exists $a \in A$ such that $f(a) = y$. Hence, $a\in f^{-1}(Y)$ implies $y = f(a) \in f(f^{-1}(Y))$. This shows $Y\subseteq f(f^{-1}(Y))$.