problem 2 let f:X$\rightarrow$Y be onto . For each b$\in$Y, let $A_b$=f^-1[{b}]. Prove that { $A_b$:b$\in$ Y} is a partition of X
I know I need to prove that for all sets in { $A_b$:b$\in$ Y} they are not empty 2 for all sets C,D$\in${ $A_b$:b$\in$ Y}, C=D or C$\cap$D=$\emptyset$ and for allx$\in$X there exists a set E$\in${ $A_b$:b$\in$ Y} such that x$\in$E how do I prove this
The relation $x\sim x' \stackrel{(def.)}{\iff} f(x)=f(x')$ is an equivalence relation in $X$, and the equivalence class of $x\in X$ is given by $[x]_\sim=\{x'\in X\mid f(x')=f(x)\}= A_{f(x)}$. Since $f$ is surjective, $\forall b\in Y, \exists x\in X$ such that $b=f(x)$, and hence $\{A_b, b\in Y\}=\{A_{f(x)}, x\in X\}=\{[x]_\sim, x\in X\}=X/\sim$.