Prove that equivalence classes are the fibers of $f$ Let $f:A\rightarrow B$ be a surjective map of two non-empty sets $A$ and $B$. We define $a\sim b$ if $f(a)=f(b)$. Prove that this is an equivalence relation with the fibers of $f$ as its equivalence classes. 
Here is my attempt at the proof: 
Since $f$ is a function, $\forall a\in A$, $f(a)=f(a)$, and therefore we have reflexivity. Given $a,b\in{A}$, if $a\sim b$, $f(a)=f(b)$, meaning $b\sim a$, thus we have symmetry. Given $a,b,c\in{A}$, if $a\sim b$ and $b\sim c$, $f(a)=f(b)$, and $f(b)=f(c)$, so $f(a)=f(c)$ then $a\sim c$. Therefore, we have transitivity. The fiber of $f$ is defined as $f^{-1}(b)=\lbrace{a\in A|\space{f(a)=b}}\rbrace$, for some $b\in B$. An equivalence class is defined as a subset of the form $\lbrace{a_1\in A|\space a_1\sim a_2}\rbrace$, for some $a_1, a_2\in A$. The fiber of $f$ contains all $a_i\in A$ such that $f(a_i)=b$, thus are equivalence classes of similarity. 
I want to know if there are
$1)$ Any corrections to be made, and 
$2)$ Better wording for the proof. 
Thank you all in advance. 
 A: The first part seems fine. I would write somewhere how you are really using the fact that "$=$" is an equivalence relation to obtain each of the steps. Also it would be better to explicitly show that, for instance, you are making the jump from $f(a)=f(b)$ and $f(b)=f(c)$ to $f(a)=f(c)$ before concluding $a\sim c$.
For the second part, it's not clear what you are saying. Write down what you intend to prove. Also, did you use surjectivity? If not, then why is that a hypothesis?
You are asked to show that the fibers of $f$ are the equivalence classes of this relation. This means we need to prove that each fiber of $f$ is an equivalence class, and each equivalence class is a fiber.
Let $b\in B$ and consider the fiber $f^{-1}(b)$. We want to show that this is the equivalence class of some element of $A$. But which element? Details in spoiler:

 Since $f$ is onto, there exists $a\in A$ such that $f(a)=b$. Then the equivalence class $[a]$ of $a$ equals $f^{-1}(b)$. Indeed, we have $c\in [a]$ iff $c\sim a$ iff $f(c)=f(a)$ iff $f(c)=b$ iff $c\in f^{-1}(b)$.

Conversely, let $a\in A$ and consider the equivalence class $[a]$ of $a$. We want to show that this is the fiber of some element in $B$ under $f$. But which element? Details in spoiler:

 We claim that $[a]$ equals the fiber $f^{-1}(f(a))$ of $f(a)$ under $f$. This follows because $c\in [a]$ iff $c\sim a$ iff $f(c)=f(a)$ iff $c\in f^{-1}(f(a))$.

