THEOREM: If $ \text{ran } F \subseteq \text{dom } G $ then $\text{dom }(G \circ F)= \text{dom }F$
PROOF: if $ F\subseteq A \times B$ and $ G\subseteq B\times C$
then by definition
$$\text{dom }F=A \\\text{ ran }F=B$$ $$\text{dom }G=B\\\text{ran }G=C$$
Now $(G \circ F)\subseteq A \times C$ by definition
so $\text{dom}(G \circ F)= A =\text{dom}F$
$ QED $
$\endgroup$ 51 Answer
$\begingroup$The fact that a function is a subset of $A\times B$ is no guarantee that the range of the function is $A$, so you can’t infer from $G\circ F\subseteq A\times C$ that $\operatorname{dom}(G\circ F)=A$. You also assumed more than you were given when you made $\operatorname{dom}G=\operatorname{ran}F$: all that you’re allowed to assume is that $\operatorname{dom}G\subseteq\operatorname{ran}F$.
It isn’t really necessary, or even all that helpful in this case, to introduce new names for the domains and ranges of $F$ and $G$; using the $\text{dom}$ and $\text{ran}$ notations explicitly makes it easier to keep in mind just what set we’re talking about at any given moment, so that’s what I’ll do.
By definition $$G\circ F=\Big\{\langle a,c\rangle\in\operatorname{dom}F\times\operatorname{ran}G:\langle a,b\rangle\in F\text{ and }\langle b,c\rangle\in G\text{ for some }b\Big\}\;.$$
From this definition it’s immediately clear that $\operatorname{dom}(G\circ F)\subseteq\operatorname{dom}F$, so we need only show that $\operatorname{dom}F\subseteq\operatorname{dom}(G\circ F)$. This is an easy bit of ‘element-chasing’: start with an arbitrary $a\in\operatorname{dom}F$, and show that it belongs to $\operatorname{dom}(G\circ F)$.
So let $a\in\operatorname{dom}F$ be arbitrary. Then there is a $b\in\operatorname{ran}F$ such that $\langle a,b\rangle\in F$. But $\operatorname{ran}F\subseteq\operatorname{dom}G$ by hypothesis, so $b\in\operatorname{dom}G$, and therefore there is a $c\in\operatorname{ran}G$ such that $\langle b,c\rangle\in G$. Thus, there is a $b$ such that $\langle a,b\rangle\in F$ and $\langle b,c\rangle\in G$, so $\langle a,c\rangle\in G\circ F$, and hence $a\in\operatorname{dom}(G\circ F)$ by definition. This shows that $\operatorname{dom}F\subseteq\operatorname{dom}(G\circ F)$, and since we already knew that $\operatorname{dom}(G\circ F)\subseteq\operatorname{dom}F$, we’ve shown that $\operatorname{dom}(G\circ F)=\operatorname{dom}F$.
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