Studying with the book Introduction to Set Theory by Karel Hrbacek, I encountered this proposition:
If $S$ is a partition of $A$ and $E_S$ is the corresponding equivalence, then $A/E_S = S$.
Where
$$E_S=\{\left \langle a,b \right \rangle \in A \times A : (\exists C \in S)(a,b \in C)\}$$ $$A/E_S = \{[a]_{E_S} : a \in A \} $$
I tried to do the proof, but would like to know if the first part is correctly done. Also, regarding the second part, I wonder if it is possible to conclude $X=C$. I have thought of something different like: ... by reflexivity we know that $a \space E_S \space a$ iff $a \in X$, so it must follow that $X \in S$. But I have doubts whether or not this is a correct reasoning. Any help is appreciated.
Proof. Assume $S$ is a partition of $A$ and $E_S$ is its corresponding equivalence.
($\subseteq$). Let $X \in S$. Then $X$ is a non-empty subset of $A$, so $a \in X$ for some $a \in A$. Hence $a \space E_S \space a$, so it follows that $a \in [a]_{E_S}$, which means $X \subseteq [a]_{E_S}$. Therefore $X$ is an equivalence class modulo $E_S$. Consequently $X \in A/E_S$.
($\supseteq $). Let $X \in A/E_S$. Then $X$ is an equivalence class modulo $E_S$. Hence $X=[a]_{E_S}$ for some $a \in A$, by reflexivity we know that $a \space E_S \space a$, so $a \in X$. But also $a \space E_S \space a$, implies there exists some $C \in S$ for which $a \in C$ holds. ...
[Update: nullified but good as a side note?]
Alternatively, one could show the map: $$f: S\to A/E_S;$$ $$X\mapsto [x]_{E_S}$$ for $x\in X\in S$ is well-defined and bijective.
$\underline{\text{Well-Definition:}}$
If $x_1,x_2\in X$, it should be clear that $[x_1]_{E_S}=[x_2]_{E_S}$.
$\underline{\text{Injective:}}$
Suppose for $X,Y\in S$ that $f(X)=f(Y).$ Then we have for representative elements $x\in X$ and $y\in Y$ that $[x]_{E_S} = [y]_{E_S}$. Hence $x E_S y$ and $\exists C\in S$ containing both $x,y$. Arbitrariness yields $X,Y\subseteq C$. But this contradicts $S$ being a partition, unless $X=C=Y$. Recall a partition of $A$ consists of mutually disjoint, non-empty subsets of $A$. Basically $C$ is not disjoint from $X$ and likewise for $Y$, so they must be the same.
$\underline{\text{Surjective:}}$
Let $[a]_{E_S}\in A/E_S$ be given. Since $A= \coprod\limits_{X\in S}X$, we have $\exists X'\in S$ containing $a$. So that $[a]_{E_S}=f(X')$.
Therefore $S \cong A/E_S$ as Sets.