elementary set theory - Confusion between an element and its preimage

Let $X$ be a set and $\sim$ is an equivalence relation on $X$, so that the quotient set
$X/_\sim=\bigcup_{x\in X}{[x]}$ with $[x]=[y]$ if and only if $x\sim y$.

Consider the quotient map $f:X\to X/_\sim\; ; \; x\mapsto [x]$.

Now take an element $[x]\in X/_\sim$ , its preimage is the set
$f^{-1}(\{[x]\})=\{z\in X\;|\; [z]= [x]\}=\{z\in X\;|\; z\sim x\}$

I'm confused because the equivalence class $[x]$ of an element $x\in X$ is defined as the set of elements $z\in X$ such that $z\sim x$ and this is the same as the preimage of $[x]$ so that it is like we have $f(x)=f^{-1}(\{f(x)\})$ !! I'm confused between $[x]$ and its preimage, which one is the equivalence class of $x$ and why they seem to be equal?

Answer

The first, and perhaps deepest problem is on the first line.

$X/{\sim}$ is not the union of the equivalence classes. It is the set of equivalence classes. More specifically, $\bigcup_{x\in X}[x]=X$, whereas $X/{\sim}=\{[x]\mid x\in X\}$. It is a set whose elements are subsets of $X$.

So a function from $X$ to $X/{\sim}$ is a function mapping points of $X$ to subsets of $X$. Therefore the preimage of an element in the range is a subset of $X$, but what is an element of the range? It is a subset of $X$ again.

And by the definition of the quotient mapping, it ends up that the preimage of $[x]$, as an element of the range, is exactly $[x]$ as a subset of the domain.