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Q.
If $R \subseteq A \times B$ and $S \subseteq B \times C$ be two relations, then
$(\text { SoR })^{-1}$ is equal to
Relations and Functions - Part 2
Solution:
$SoR$ is a relation from $A$ to $C$.
$\therefore ( SoR )^{-1}$ is a relation from $C$ to $A$.
$R ^{-1}$ is a relation from $B$ to $A$.
$S ^{-1}$ is a relation from $C$ to $B$.
$\therefore R ^{-1} oS ^{-1}$ is a relation from $C$ to $A$.
Let $( c , a ) \in( SoR )^{-1} . $
$\therefore ( a , c ) \in SoR$
$\therefore \exists b \in B :( a , b ) \in R$ and $( b , c ) \in S$
$\therefore ( b , a ) \in R ^{-1}$ and $( c , b ) \in S ^{-1}$
$\therefore ( c , a ) \in R ^{-1} oS ^{-1}$
$ \therefore ( SoR )^{-1} \subseteq R ^{-1} oS ^{-1}$
Conversely, let $( c , a ) \in R ^{-1} oS ^{-1}$
$\therefore \exists b \in B :( c , b ) \in S ^{-1}$ and $( b , a ) \in R ^{-1}$
$\therefore ( b , c ) \in S$ and $( a , b ) \in R$
$\Rightarrow ( a , c ) \in SoR$
$ \Rightarrow ( c , a ) \in( SoR )^{-1}$
$\therefore R ^{-1} oS ^{-1} \subseteq( SoR )^{-1}$
Combining, we get $( SoR )^{-1}= R ^{-1} oS ^{-1}$