Equivalence classes/Partition/Quotient set/Properties/Fact/Proof

Proof

Let ${}x$ and ${}y$ be equivalent, and ${}u\in [x]$ . Then ${}x\sim u$ , and, by transitivity, also ${}y\sim u$ , thus ${}u\in [y]$ . Therefore, the equivalence classes coincide. The implication from the middle to the right is clear, because, due to ${}x\sim x$ , the equivalence classes are not empty. Suppose now that ${}[x]\cap [y]\neq \emptyset$ , and let ${}z$ denote an element in the intersection. Then ${}x\sim z$ and ${}y\sim z$ , and, by transitivity, ${}x\sim y$ .
Because of the reflexivity, we have ${}x\in [x]$ ; therefore, ${}M=\bigcup _{[x]\in M/\sim }[x]$ . This union is disjoint by part (1).
The surjectivity is clear because of the definition of the quotient set, and since ${}x$ is sent to the class ${}[x]$ .
We have
${}q^{-1}([x])={\left\{y\in M\mid q(y)=[x]\right\}}={\left\{y\in M\mid [y]=[x]\right\}}={\left\{y\in M\mid y\sim x\right\}}=[x]\,.$