Proof that if $R_{1};R_{2}=R_{2};R_{1}$ is a relation on a set $S$, then $R_{1}$ and $R_{2}$ are bijective maps

Question

For the surjectivity part, I believe I have managed to prove the claim. And for Injectivity, I've tried to show that if $aR_{1};R_{2}c$ and $bR_{2};R_{1}c$, then $a=b$ but I am not entirely sure what $a=b$ means in this context. Do I have to show that there is an equivalence relation containing (a,b)?

Answer 1

Something is missing in the question. Take $S=\{a,b\}$ and take $R_1$ and $R_2$ to be the relations defined by the function $f:x\mapsto a$. Then, $R_1;R_2=f\circ f=R_2;R_1$, but $f$ is not injective nor surjective.