Solution
(related to Problem: To Construct a Partition of a Given Set)
Creating Partitions Using Fibers of Surjective Functions
 Let $f:X\to Y$ be a given surjective function.
 By definition of surjective functions, every element $y\in Y$ has an $x\in X$ with $f(x)=y.$
 In general, there can be many such elements $x\in X,$ with $f(x)=y,$ called the fiber $f^{1}(y):=\{x\in X\mid f(x)=y,\;y\in Y\}.$
 Since $f$ is a function, for any two $y_1,y_2\in Y,$ the fibers $f^{1}(y_1)$ and $f^{1}(y_2)$ are disjoint in $X.$
 Therefore, the all fibers $f^{1}(y),$ $y\in Y$ arbitrary, are mutually disjoint in $X.$
 This means that the set of all fibers $$\{f^{1}(y)\mid y\in Y\}$$ is a set partition of $X.$
Creating Partitions Using Equivalence Relations
 Let $R\subseteq X\times X$ be a given equivalence relation.
 Then, for every element $x\in X,$ we have an equivalence class $[x]:=\{z\in X, zRx\}.$
 Now, any two equivalence classes $[x]\neq [y]$ are disjoint in $X,$ because otherwise,
 for an element $z\in [x]\cap [y]$ we would have $zRx$ and $zRy.$
 Because $R$ is an equivalence relation, it is transitive and symmetric.
 Therefore, from $zRx$ and $zRy$ it would follow that $xRy$, in contradiction to $[x]\neq [y].$
 Therefore, all equvalence classes $[x]$, $x\in X$ are mutually disjoint.
 This means that the quotient set of all equivalence classes $$\{[x]\mid x\in X\}$$ is a set partition of $X.$
Differences between the two possibilities
A partition built by a given equivalence relation can itself be reversely used to define this equivalence relation, i.e. if we are given partition, then its elements can be reinterpreted as equivalence classes of an equivalence relation. In other words, partitions and equivalence relations are mathematical concepts, which are very closely related to each other.
This onetoone correspondence between partitions and equivalence relations cannot be observed for partitions built by means of fibers of a given surjective function. For a given partition, it is usually not possible to find an appropriate surjective function. The reason for this is simple. A given partition of a set $X$ is independent of another set $Y$, which we need to define a function between $X$ and $Y.$
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References
Bibliography
 Flachsmeyer, Jürgen: "Kombinatorik", VEB Deutscher Verlag der Wissenschaften, 1972, 3rd Edition