Exercise 0.1.7 (Equivalence classes and fibers)

• Since $f(a)=f(a)$, $a\sim a$ for all $a\in A$: the relation $\sim$ is reflexive.
• If $a\sim b$, where $a,b\in A$, then $f(a)=f(b)$, thus $f(b)=f(a)$, so $b\sim a$. The relation $\sim$ is symmetric.
• If $a\sim b$ and $b\sim c$, then $f(a)=f(b)$ and $f(b)=f(c)$, thus $f(a)=f(c)$, so $a\sim c$. The relation $\sim$ is transitive.

This shows that $\sim$ is an equivalence relation.

Let $U\subset A$ an equivalence class for $\sim$. By definition, there is some $a\in A$ such that

Define $b=f(a)$. Then

Therefore every class $U$ is a fiber.

Conversely, let $b$ be any element in $B$, and $f^{-1}(\{b\})$ the corresponding fiber. Since $f$ is surjective, there is some $a\in A$ such that $f(a)=b$, and the preceding equalities show that $f^{-1}(\{b\})=ȧ$ is the class of $a$. Every fiber is an equivalence class.

To conclude, the equivalence classes for $\sim$ are the fibers of $f$.