given a function $f$, as $f: C \rightarrow P(A)$ as $f\left(R\right)= \left[a\right]_{R}$ does $f$ have an inverse?

Question

Given a set $A$, let $C$ be the set of equivalence relations over $A$.

Given $a\in A$, we shall define a function, as $f: C \rightarrow P(A)$ by $f\left(R\right)= \left[a\right]_{R}$.

I need to prove or disprove that $f$ is onto and one-to-one, and to find its inverse (if such one exists).

I don't understand the properties of this function and therefore stuck.

Answer 1

Note that every equivalence relation on A defines a partition of A into disjoint equivalence classes.

Also every partition of A into disjoint subsets, defines an equivalence relation.

Given a fixed $a\in A$, the function $$ f: C \rightarrow P(A)$$ defined by

$$f\left(R\right)= \left[a\right]_{R}$$

takes you relation $R$ to the class of $[a]_R$

This function is not onto P(A) because we can consider a subset of A which does not include $a$, then that subset will not be $[a]_R$ with any equivalence relation R.

Also it is not one-to-one, because given a subset $B\subset A$ we can define two different partition on A which share the same component $B$