Symmetricity of composition of equivalence relations

Question

I stumbled upon this question:

Let $A$ be a set and $T$ and $S$ two equivalence relations on $A$.

Prove: $S\circ T=T\circ S \iff S\circ T$ is an equivalence relation

In these questions

equivalence relation composition problem

Composition of equivalence relations

the OPs asked for the proof of transitivity of composition of equivalence relations.

My question is how to prove the symmetricity?

If we assume $(x,y) \in ST $, we need to prove $(y,x) \in ST$.

By definition, there exists $h$ such that $(x,h) \in T$ and $(h,y) \in S$.

Getting anywhere from there seems to be a mystery for me.

Appreciate any help.

Answer 1

In the sequel $R^{\text{op}}:=\{\langle x,y\rangle\mid\langle y,x\rangle\in R\}$ for any relation $R$.

If $S$ and $T$ are both symmetric then actually we have:$$S^{\text{op}}=S\text{ and }T^{\text{op}}=T$$

Then consequently:$$\left(S\circ T\right)^{\text{op}}=T^{\text{op}}\circ S^{\text{op}}=T\circ S$$

Now if we also have $T\circ S=S\circ T$ then this results in:$$\left(S\circ T\right)^{\text{op}}=S\circ T$$ This states exactly that $S\circ T$ is symmetric.