Please excuse my English if it's not understandable, exercises are translated so I don't know all the English terms in math. So I'm doing exercises regarding relations and compositions, and one of the exercises is:
Suppose $R\subseteq A\times B$, $S\subseteq B\times C$ and $T\subseteq C\times D$. Which of the following statements are true?
- $(S\circ R)^{-1}=R^{-1}\circ S^{-1}$
- $S\circ R=R^{-1}\circ S^{-1}$
- $S\circ R=(R\circ S)^{-1}$
- $(T\circ S)\circ R=T\circ (S\circ R)$
The answer to 1.:
To prove 1., suppose $(c,a)\in(S\circ R)^{-1}$. Then, by definition of the inverse relation, $(a,c)\in (S\circ R)$. By the definition of the composite relation, there exists $b\in B$ such that $(a,b)\in R$ and $(b,c)\in S$. This of course means $(c,b)\in S^{-1}$ and $(b,a)\in R^{-1}$. Applying the definition of the composite relation again, we find $(c,a)\in R^{-1}\circ S^{-1}$. Thus, $(S\circ R)\subseteq R^{-1}\circ S^{-1}$. Similar arguments show $R^{-1}\circ S^{-1}\subseteq (S\circ R)$ and the result follows.
Note that this is not my answer. I just don't understand how I can come to this conclusion. Our lecturer hasn't covered how to prove these kind of calcualtions. Can someone please explain to me the approach to these kind of exercises?
Note that to show that two sets are equal (relations are sets), e.g. $A=B$, then it has to be true that $A\subseteq B$ and $B\subseteq A$. Now, if you want to show that $A\subseteq B$, you have to prove that the implication $$x\in A\implies x\in B$$ is true (similar for $B\subseteq A$).
This is what the solution of the first question does. It takes an arbitrary element $(c,a)\in(S\circ R)^{-1}$ and then concludes by properties of (inverse) relations and by using the definition of relations, that $(c,a)\in R^{-1}\circ S^{-1}$ is also true.
The basic properties to use here are: