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

Proof

1. 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 $[x]\cap [y]\ne \mathrm{\varnothing }$, and let $z$ denote an element in the intersection. Then $x\sim z$ and $y\sim z$, and, by transitivity, $x\sim y$.
2. Because of the reflexivity, we have $x\in [x]$; therefore, $M=\bigcup _{[x]\in M/\sim }[x]$. This union is disjoint by part (1).
3. The surjectivity is clear because of the definition of the quotient set, and since $x$ is sent to the class $[x]$.
4. We have

   $q^{-1}([x])=\{y\in M\mid q(y)=[x]\}=\{y\in M\mid [y]=[x]\}=\{y\in M\mid y\sim x\}=[x].$

To fact