Let $R$ be an equivalence relation on $X$ and $S$ an equivalence relation on $X/R$.....

Question

Let $R$ be an equivalence relation on $X$ and $S$ an equivalence relation on $X/R$. Find an equivalence relation $T$ on $X$ such that $(X/R)/S$ is in one-to-one correspondence with $X/T$ under the mapping $[[x]R]S\mapsto[x]T$.

Not much to add. I found this question in Elements of Abstract Algebra by Allen Clark. Any help would be appreciated. Cheers!

Answer 1

Hint

Define

$$(u,v) \in T \iff (\exists \mathcal S \in (X/R)/S)(u \in \mathcal S \wedge v \in \mathcal S)$$

Answer 2

HINT: Use the relationship between equivalence relations on a set and the associated partitions of that set. $R$ induces a partition $\mathscr{P}_R$ of $X$. $S$ then induces a partition $\mathscr{P}_S$ of $X/R$. Each member of $\mathscr{P}_S$ is a set whose elements are members of $\mathscr{P}_R$. In other words, the equivalence relation $R$ lumps the members of $X$ into sets — the $R$-equivalence classes, which are also the members of $\mathscr{P}_R$ — and the equivalence relation $S$ then lumps those sets into collections of sets that are the $S$-equivalence classes, i.e., the members of $\mathscr{P}_S$.

It’s as if you were to take a box, $X$, of objects, and begin by sorting those objects into small boxes sitting inside $X$; those small boxes are the $R$-equivalence classes, the members of the partition $\mathscr{P}_R$. Then you sort those smaller boxes into medium-sized boxes still sitting inside $X$; those medium-sized boxes are the $S$-equivalence classes, the members of the partition $\mathscr{P}_S$.

Now you want to find a single partition of $X$ whose parts are in one-to-one correspondence with $\mathscr{P}_S$. How can you sort the objects in $X$ into boxes that correspond naturally to the medium-sized boxes?