For the surjectivity part, I showed that if $(a,b)\in I$ then $(a,c)\in R_{1}$ and $(c,b)\in R_{1}$ and $(a,c)\in R_{2}$ and $(c,b)\in R_{2}$ for some $c\in S$. Now for arbituary $(x,y)\in R_{1}$ we have $(x,z)\in I$ which implies that $(y,z)\in R_{2}$. But also $(y,z)\in R_{1}$, so $R_{1}$ is surjective. Similar argument for $R_{2}$
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)?
edit: $R_{1};R_{2}$ is a composition of two relations and $I=R_{1};R_{2}$
Claim 1: $R_{1}$ and $R_{2}$ are maps.
Suppose $(a,b) \in R_{1}$. Now $(b,c) \in R_{2}$ for some $c \in S$. But then $c=a$ since $(a,c) \in R_{1} \circ R_{2}=I$. So for arbituary $x \in S$, if $(x,y) \in R_{1}$ then $(y,x) \in R_{2}$.
Now, suppose $(x,y) \in R_{1} \wedge (x,z) \in R_{1}$. Then $(y,x) \in R_{2}$ so $(y,z) \in R_{2} \circ R_{1}=I \implies y=z$
So $R_{1}$ is a map. Same line of argument for $R_{2}$
Claim 2: $R_{1}$ and $R_{2}$ are injective.
Suppose $(x,y) \in R_{1} \wedge (x^*,y) \in R_{1}$. But then $(y,x^*) \in R_{2}$ so $(x,x^*) \in R_{1} \circ R_{2} \iff x=x^*$. Same argument for $R_{2}$
Claim 3: $R_{1}$ is sujective (and hence $R_{2}$).
Let $s \in S$ arbitrary. Then $(s,l) \in R_{2} \iff (l,s) \in R_{1}$. Hence $R_{1}$ is sujective.
Above imply that both $R_{1}$ and $R_{2}$ are bijective maps