1. ## Equivalence class

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

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

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

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

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

Now I have to show that the equivalence classes are the fibers of $\displaystyle f$. So by definition, an equivalence class is $\displaystyle \{x \in A|x \sim a \}$. 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 $\displaystyle \{x \in A| f(a) = f(b) \}$?

Thanks

2. Originally Posted by shilz222
[B] Now I have to show that the equivalence classes are the fibers of $\displaystyle f$. So by definition, an equivalence class is $\displaystyle \{x \in A|x \sim a \}$. 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.