If $\operatorname{Ran} R \subset \operatorname{Dom }S$. Prove $ \operatorname{Dom} (R) \subset\operatorname{Dom}(S \circ R ) $

Question

If \operatorname{Ran} R $\subset$ \operatorname{Dom} S

Prove $ \operatorname{Dom} (R) \subset \operatorname{Dom}(S \circ R ) $

Now Range of R = $\{ \,b \in B \mid (a,b) \in R\,\} $ ...... $(1)$

And Domain of S = $\{\, b' \in B \mid (b',c) \in S \,\} $ ....... $(2)$

Since $\operatorname{Ran} (R)$ is a subset of domain of S. So it means for all $b \in \operatorname{Ran}(R)$ implies $b \in \operatorname{Dom}(S)$. So there exists $b''$ which satisfies properties of both 1 and 2 above. So choose $b''$.

Now let $a \in \operatorname{Dom}(R)$. So there exists b'' such that $(a,b'') \in R$ and $(b'',c) \in S$. So $a \in \operatorname{Dom}(S \circ R)$

Is this correct? Thanks for help.

Answer 1

Let $\mathsf{Ran}R\subseteq\mathsf{Dom}S$ and let $a\in\mathsf{Dom}R$.

Then by definition $aRb$ for some $b$ and also by definition $b\in\mathsf{Ran}R$.

Then from $\mathsf{Ran}R\subseteq\mathsf{Dom}S$ we conclude that $b\in\mathsf{Dom}S$ which means that $bSc$ for some $c$.

Then from $aRb\wedge bSc$ it follows that $a(S\circ R)c$ telling us that $a\in\mathsf{Dom}(S\circ R)$.

Proved is now that $a\in\mathsf{Dom}R\implies a\in\mathsf{Dom}(S\circ R)$ allowing the conclusion that under condition $\mathsf{Ran}R\subseteq\mathsf{Dom}S$ we have: $$\mathsf{Dom}R\subseteq\mathsf{Dom}(S\circ R)$$