Let f: A ---> B and g: B ---> C, if both f, g are bijective, show (prove) that g o f : A ---> C is bijective.
Hi
There are various ways to prove that assertion, which seems very natural.
You want $\displaystyle g\circ f$ to be injective and surjective, assuming that $\displaystyle f$ and $\displaystyle g$ have such properties.
Take two distinct elements $\displaystyle a_1$ and $\displaystyle a_2$ in $\displaystyle A.$ $\displaystyle f$ injective $\displaystyle \Rightarrow f(a_1)\neq f(a_2).$ Do you see how to the same idea to prove that $\displaystyle g\circ f$ is injective too.
$\displaystyle f\ \text{surjective}\ \Leftrightarrow\ \text{Im}f = B$ . $\displaystyle g\ \text{surjective}\ \Leftrightarrow\ \text{Im}g=C.$ (just the definitions of $\displaystyle f$ and $\displaystyle g$ surjectivity). Given a $\displaystyle c\in C,$ why is there a $\displaystyle a\in A$ such that $\displaystyle g(f(a))=c$ ?