### 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 re-interpreted 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 one-to-one 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 $Y.$

| | | | created: 2019-09-08 11:36:46 | modified: 2019-09-08 11:42:54 | by: *bookofproofs* | references: [8297]