Let $f:A\mapsto B$, and $g:B \mapsto C$ be two bijective relations. Show that $g \circ f:A \mapsto C$ is also bijective.
Proof : As I understand it, the relation $g \circ f:A \mapsto C$ takes an element from the domain of $f$ and pairs it with an element from the codomain of $g$, such that a pair $(a\in A,c \in C)$ is formed. For $g \circ f$ to be bijective, we have to show both injectivity and surjectivity. If the composite relation is injective, that means $$(a_1,c)\in g\circ f=(a_2,c)\in g\circ f \rightarrow a_1=a_2$$
must be true. Assuming the contradiction, that is $$a_1 \neq a_2$$
However, this means we have two pairs $(a_1,b)$ and $(a_2,b)$ in the set $f \subset A \times B$. This contradicts the fact that $f$ is bijective, and thus $g \circ f$ is injective.
Now, every pair $(a,c)$ in $g \circ f$ is an element of $A \times C$ only if $aRb$ and $bRc$. Since $f$ and $g$ are both bijective, this means we can always find $a \in A$ for every $c \in C$ such that $aRc$. Thus $g \circ f$ is surjective and therefore a bijective relation. What are the flaws in this proof?