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<Moderator's note: Moved from a technical forum and thus no template. Also re-edited: Please use ## instead of $$.>

If ##R_{1}## and ##R_{2}## are relations on a set S with ##R_{1};R_{2}=I=R_{2};R_{1}##. Then ##R_{1}## and ##R_{2}## are bijective maps

##R_{1};R_{2}## is a composition of two relations

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.

If ##R_{1}## and ##R_{2}## are relations on a set S with ##R_{1};R_{2}=I=R_{2};R_{1}##. Then ##R_{1}## and ##R_{2}## are bijective maps

##R_{1};R_{2}## is a composition of two relations

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.

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