let f:A ---> B and g:B ---> C (if both are bijective, prove g o f is bijective)

Hi

There are various ways to prove that assertion, which seems very natural.

You want $g\circ f$ to be injective and surjective, assuming that $f$ and $g$ have such properties.

Take two distinct elements $a_1$ and $a_2$ in $A.$ $f$ injective $\Rightarrow f(a_1)\neq f(a_2).$ Do you see how to the same idea to prove that $g\circ f$ is injective too.

$f\ \text{surjective}\ \Leftrightarrow\ \text{Im}f = B$ . $g\ \text{surjective}\ \Leftrightarrow\ \text{Im}g=C.$ (just the definitions of $f$ and $g$ surjectivity). Given a $c\in C,$ why is there a $a\in A$ such that $g(f(a))=c$ ?