I am trying to solve the following problem:
Let A and B be sets and let f:A→B be a surjective function. For each b∈ B, let $A_b=(f^{-1}) (\{b\})$. Prove that the collection of sets $\{A_b | b\in B\}$ is a partition of A.
I understand that in order for something to be a partition of a set, it must be pairwise disjoint and that $\cup A_b=A$ is also a requirement. I also know that since the function is surjective, A has more elements than B and $f(A)=B$. However, I'm having trouble figuring out how to form this information into a cohesive proof.
Given $b,b'\in B$, $b\neq b'$, suppose that there exists $x\in A_b\cap A_{b'}$. Then : $b=f(x)=b'$, a contradiction ! Hence $A_b\cap A_{b'}=\emptyset$.
For any $b\in B$, there exists $a\in A$ such that $f(a)=b$ because $f$ is surjective. Then $a\in (f^{-1})(\{b\})$ and we conclude that $A_b\neq\emptyset$.
Finally $A\subset\bigcup_{b\in B}A_b$ because each $x\in A$ belongs to $A_{f(x)}$. And the converse inclusion is trivial.