I'm pretty new to mathematical proof and set theory and I'm having trouble proving this problem. I've started on it, but I don't know where to progress.
Problem: Suppose that $f: A \rightarrow B$. Define a relation $R$ on $A$ by $xRy$ iff $f(x) = f(y)$. For any $x \in A,$ let $E_x$ be the equivalence class of $x$. Additionally, let $E$ be the collection of all equivalence classes.
Prove that the function $g: A \rightarrow E$ defined by $g(x) = E_x$ is surjective.
Proof: To show that $g$ is surjective, we must show that every $y \in E$ can be expressed as $g(x)$ for some $x \in A$.
Let $y \in E$. Since $E = \{E_x: x \in A \}$, $y \in E_x$ for some $x \in A$. Let this particular $x$ be $x_0$, so $y \in E_{x_0}$. Now, since $E_{x_0} = \{y \in A: yRx_0 \}$, $y \in A$ as well.
As $yRx_0$, this means $f(x_0) = f(y)$.
I do not know where to go after this. Have I started this correctly, and how should I finish?
You're close but no! $E$ is a set of sets so once you say $y\in E$ you're saying $y$ is one of the equivalence classes, i.e. $y=E_x$ for some $x$ (rather than $y\in E_x$).