On the fundamental theorem of equivalence relations Here is a what I did which is the same as here, but with more details:
Given a set $X$ with an equivalence relation ${\sim}$

There exist a bijection between the set of all possible partitions of $X$ and a set of all equivalence relations on $X$:

The relation between these two sets is $\color{blue}{\text{left-total}}$ since for every quotient set $X/{\sim}$ there exist a partition containing all equivalence classes of $X$ by ${\sim}$.
Also the relation is $\color{blue}{\text{functional}}$ and  $\color{green}{\text{injective}}$ since for every $X/{\sim}$ there exist just a unique partition containing all elements of $X/{\sim}$ (because if not , then there exist a $X/{\sim}$ which is mapped to more than one partition, so the partitions should be equal but there does not exist any two partitions with the same elements, a contradiction) and the same can be said for two quotient set of $X$ mapping to a single partition  .
Since each partition is identical to one $X/{\sim}$ therefore the relation between the two sets is $\color{green}{\text{surjective}}$ .
Finally we've shown that the relation between these two sets is a $\color{green}{\text{bijective}}\: \color{blue}{\text{function}}$
Is my proof right?
 A: Your proof is basically fine, but the correspondence should be explicitly given:
$$\Phi(\sim) :=\{\{a:a\sim x\} \ :\, x\in X\} \\
\Psi(\mathcal S) :=\{(x,y) :\exists S\in\mathcal S:\, x,y\in S\}$$
Once these are given, (after verifying that $\Phi(\sim)$ is a partition if $\sim$ is an equivalence relation and that $\Psi(\mathcal S)$ is an equivalence relation if $\mathcal S$ is a partition), it's easier to simply check that
$$\Psi(\Phi(\sim)) =\sim \\
\Phi(\Psi(\mathcal S)) =\mathcal S\,. $$
A: Let $X$ be a set.
Let $\mathcal P$ denote the collection of all partitions on $X$.
Let $\mathcal E$ denote the collection of all equivalence relations on $X$.
Then you assert that a bijection $\mathcal P\to\mathcal E$ exists.
You start with a relation which is in this context a subset of $\mathcal P\times\mathcal E$ but do not really mention which relation you have in mind here.
If $R$ denotes the relation then for proving that it is a bijection it is indeed necessary and sufficient to show that:


*

*$R$ is left-total, i.e. for every $P\in\mathcal P$ there is at least one $E\in\mathcal E$ such that $(P,E)\in R$.

*$R$ is functional, i.e. for every $P\in\mathcal P$ there is at most one $E\in\mathcal E$ such that $(P,E)\in R$.

*$R$ is injective, i.e. for every $E\in\mathcal E$ there is at most one $P\in\mathcal P$ such that $(P,E)\in R$.

*$R$ is surjective, i.e. for every $E\in\mathcal E$ there is at least one $P\in\mathcal P$ such that $(P,E)\in R$.
A relation having these properties is exactly a bijection.
The only thing missing is an explicit definition this relation.
This can be repaired by stating that: $$(P,E)\in R\iff P=\{[x]_E\mid x\in X\}$$where $[x]_E$ denotes the equivalence class wrt $E$ that is represented by element $x\in X$.
Or alternatively by stating that:$$(P,E)\in R\iff E=\{(x,y)\in X^2\mid\exists p\in P[x,y\in p]\}$$
Then the bullet points can be proved one by one.
IMV however it is more handsome to define functions $f:\mathcal P\to\mathcal E$ and $g:\mathcal E\to\mathcal P$ and to prove that $g\circ f$ and $f\circ g$ are both identities. This assures that $f$ and $g$ are bijections.
Let $f$ be prescribed by:  $P\mapsto\{(x,y)\in X^2\mid\exists p\in P[x,y\in p]\}$.
Let $g$ be prescribed by: $E\mapsto \{[x]_E\mid x\in X\}$.
It is not difficult to prove that $f(P)\in\mathcal E$ for every $P\in\mathcal P$ and $g(E)\in\mathcal P$ for every $E\in\mathcal E$ showing that we are indeed dealing with functions $f:\mathcal P\to\mathcal E$ and $g:\mathcal E\to\mathcal P$.
Also it is not difficult to prove that $f\circ g$ and $g\circ f$ are identities.
