Given:

$f:\mathrm{A}\to \mathrm{B}$ and $\mathrm{g}:\mathrm{B}\to \mathrm{C}$

And, $gof:\mathrm{A}\to \mathrm{C}$ is a bijection.

Let $x, y\in A$ such that $f\left(x\right)=f\left(y\right)$.

Then,

$f\left(x\right)=f\left(y\right)$

$\Rightarrow g\left(f\left(x\right)\right)=g\left(f\left(y\right)\right)$

$\Rightarrow gof\left(x\right)=gof\left(y\right)$

$\Rightarrow x=y; \left[\because gof:A\to C\text{ is one-one}\right]$

Hence, $f\left(x\right)$ is an injection.

Now,

Let $z$ be an arbitrary element of $C$.

Since, $gof:A\to C$ is onto.

Therefore, there exists $x\in A$ such that

$gof\left(x\right)=z$

$\Rightarrow g\left(f\left(x\right)\right)=z$

$\Rightarrow g\left(y\right)=z$ where $y=f\left(x\right)\in B$.

Thus, for all $z\in C$ there exists $y=f\left(x\right)\in B$ such that $g\left(y\right)=z$.

Hence, $g:B\to C$ is surjection.