
Proof trouble
Hi, I'm working out of Dummit and Foote and I need to know if I'm proceeding correctly in a proof. Problem: for the surjective map from A to B f, prove that the relation
a ~ b iff f(a)=f(b)
is an equivalence relation whose equivalence classes are the fibers of f.
So it makes sense but I don't know if I'm necessarily 'showing it'. I say:
if f(a) = f (b)
then a= b
so (a ~ b ) > a=b
for any f(x) in y there exists a set containing only f(x). the preimage of any of these sets defines the fiber of f over their respective elements...
Long story short, I'm running in circles and would appreciate a helping hand. Thanks!

You defined a relation ~ by $\displaystyle a \sim b \Leftrightarrow f(a)=f(b)$ where $\displaystyle f:A \to B$ is a surjection and $\displaystyle a,b \in A$.
1) Reflexivity:
$\displaystyle a \sim a \Leftrightarrow f(a)=f(a)$
2)Symmetry:
$\displaystyle a \sim b \Rightarrow f(a)=f(b) \Rightarrow f(b)=f(a)$
3) Transitivity:
$\displaystyle a \sim b, b \sim c \Rightarrow f(a)=f(b)=f(c)$
Also, f is not neccessarily injective  so $\displaystyle f(a)=f(b) \Rightarrow a=b$ is not always true.
The second part is straightforward  take a look at the definition of a fiber and you should see why every equivalence class of ~ is a fiber.