Consider function $$f : A \rightarrow B$$ and $$g : B \rightarrow C$$ $$(A, B, C \subseteq R)$$ such that $$(gof)^{-1}$$ exists, then:

We begin with the information that $$f:A\rightarrow B$$ and $$g:B\rightarrow C$$ are two real-valued functions such that the composite function $$g\circ f:A\rightarrow C$$ possesses an inverse, written $$(g\circ f)^{-1}$$.

A function admits an inverse if and only if it is bijective; that is, it must be simultaneously one-one (injective) and onto (surjective). Hence we may immediately state

$$g\circ f \text{ is one-one and onto.}$$

We now translate the bijectivity of $$g\circ f$$ into separate conditions on $$f$$ and $$g$$.

Injectivity part: To prove that $$f$$ is one-one, let us take any two elements $$x_{1},x_{2}\in A$$ and assume that their images under $$f$$ coincide, i.e.

$$f(x_{1})=f(x_{2}).$$

Applying $$g$$ on both sides we obtain

$$g\!\left(f(x_{1})\right)=g\!\left(f(x_{2})\right) \;\Longrightarrow\; (g\circ f)(x_{1})=(g\circ f)(x_{2}).$$

Since $$g\circ f$$ is injective, equality of the outputs forces equality of the inputs, so

$$x_{1}=x_{2}.$$

This derivation shows that no two distinct elements of $$A$$ can share the same image under $$f$$; therefore $$f$$ is one-one (injective).

Surjectivity part: To show that $$g$$ is onto, start with an arbitrary element $$c\in C$$. Because $$g\circ f$$ is surjective, there exists some $$a\in A$$ satisfying

$$(g\circ f)(a)=c.$$

Writing the composition explicitly, this reads

$$g\!\bigl(f(a)\bigr)=c.$$

Denote $$b=f(a)\in B$$; then the equation becomes $$g(b)=c$$. We have thus produced an element $$b\in B$$ whose image under $$g$$ is the pre-selected $$c$$, verifying that every element of $$C$$ is attained by $$g$$. Consequently, $$g$$ is onto (surjective).

Putting the two deductions together, we have established that

$$f \text{ is one-one and } g \text{ is onto.}$$

None of the other combinations is compelled by the mere bijectivity of $$g\circ f$$. Hence, the correct answer is Option C.