We are given: $f:A\to B, \qquad g:B\to C$ and the composition $(g\circ f):A\to C$ has an inverse, i.e. $(g\circ f)^{-1}$ exists.

A function has an inverse iff it is bijective. Therefore, $g\circ f$ is both:

• one-one (injective), and
• onto (surjective).

First, prove that $f$ is one-one.

Suppose $f(x_1)=f(x_2).$ Applying $g$ on both sides, $g(f(x_1))=g(f(x_2))$ $\Rightarrow (g\circ f)(x_1)=(g\circ f)(x_2).$ Since $g\circ f$ is one-one, we must have $x_1=x_2.$ Hence $f$ is one-one.

Next, prove that $g$ is onto.

Since $g\circ f$ is onto from $A$ to $C$, for every $c\in C$, there exists some $a\in A$ such that $(g\circ f)(a)=c.$ Let $b=f(a)\in B.$ Then $g(b)=c.$ So for every $c\in C$, there exists $b\in B$ with $g(b)=c$. Therefore, $g$ is onto.

Check the other possibilities.

• A: $f$ and $g$ both are one-one Not necessary. $g$ need not be one-one on all of $B$; it only has to be onto, and its restriction to $f(A)$ behaves injectively enough for the composition.

• B: $f$ and $g$ both are onto Not necessary. $f$ need not be onto $B$.

• C: $f$ is one-one and $g$ is onto This is always true.

• D: $f$ is onto and $g$ is one-one Not necessary.

Hence the correct option is $\boxed{\text{C}}.$