1. Equivalence class

Problem: Let $f: A \to B$ be a surjective map of sets. Prove that the relation $a \sim b \Leftrightarrow f(a) = f(b)$ is an equivalence relation whose equivalence classes are the fibers of $f$.

My Work/Thoughts: What does a surjective map of sets mean? I think it means that for all $b \in B$ there is some $a \in A$ such that $f(a) = b$. An equivalence relation has to be symmetric, reflexive, and transitive. I know that $a \sim b$ really means that $(a,b) \in R$ which is some subset of a set $A \times A$. So we have to show that these 3 properties hold.

(i) reflexive: $a \sim a$ for all $a \in A$. So $f(a) = f(a)$?

(ii) symmetric: $a \sim b \Rightarrow b \sim a$ for all $a,b \in A$. So $f(a) = f(b)$ and $f(b) = f(a)$.

(iii) transitive: $a \sim b$ and $b \sim c \Rightarrow a \sim c$ for all $a,b,c \in A$. So $f(a) = f(b) \ \text{and} \ f(b) = f(c) \Rightarrow f(a) = f(c)$ for some $a,b,c \in A$? So the ' $=$' sign acts like the ' $\sim$'?

Now I have to show that the equivalence classes are the fibers of $f$. So by definition, an equivalence class is $\{x \in A|x \sim a \}$. This is where I become stuck. A fiber is the inverse image of one element of the codomain mapped to the domain right? So is it $\{x \in A| f(a) = f(b) \}$?

Thanks

2. Originally Posted by shilz222
[B] Now I have to show that the equivalence classes are the fibers of $f$. So by definition, an equivalence class is $\{x \in A|x \sim a \}$. This is where I become stuck. A fiber is the inverse image of one element of the codomain mapped to the domain right?
You work is just fine. However, this particular problem requires the function to be a surjection.
In general, the fibers of a function define an equivalence relation on its domain. But the that may vary with your particular definitions.