Let f: A → B and g : B → C be the bijective functions. Then (g o f)-1 is,

The correct option is f⁻¹ o g⁻¹.

To understand why this is correct, let us analyze the properties of bijective functions and their compositions step-by-step.

Step 1: Understand the Functions and Their Inverses
We are given two bijective (one-to-one and onto) functions:
$$f:A\to B$$
and
$$g:B\to C$$
Because both $f$ and $g$ are bijective, their inverse functions exist and are also bijective:
$$f^{-1}:B\to A$$
and
$$g^{-1}:C\to B$$

Step 2: Map the Composition and its Inverse
The composition function $g\circ f$ maps elements from set $A$ to set $C$:
$$(g\circ f):A\to C$$
The inverse of this composition, ${(g\circ f)}^{-1}$, must reverse this direction and map elements from set $C$ back to set $A$:
$${(g\circ f)}^{-1}:C\to A$$

Step 3: Track the Elements through the Mapping
Let us select an arbitrary element $x\in A$. Under the function $f$, this maps to an element $y\in B$:
$$f\left(x\right)=y \Rightarrow f^{-1}\left(y\right)=x$$
Next, under the function $g$, the element $y\in B$ maps to an element $z\in C$:
$$g\left(y\right)=z \Rightarrow g^{-1}\left(z\right)=y$$

Step 4: Express the Inverse Composition
Using the mappings above, the composite function evaluated at $x$ yields:
$$(g\circ f)\left(x\right)=g\left(f\right(x\left)\right)=g\left(y\right)=z$$
By the definition of an inverse function, applying ${(g\circ f)}^{-1}$ to $z$ must return us to $x$:
$${(g\circ f)}^{-1}\left(z\right)=x$$

Step 5: Apply the Inverses Separately
Now let us trace the path backwards from $z$ using the individual inverse functions. Since we want to map $C\to A$, we must first apply $g^{-1}$ to map $C\to B$, and then apply $f^{-1}$ to map $B\to A$:
$$(f^{-1}\circ g^{-1})\left(z\right)=f^{-1}\left(g^{-1}\right(z\left)\right)$$
Substituting $g^{-1}\left(z\right)=y$ into the equation gives:
$$f^{-1}\left(y\right)$$
Substituting $f^{-1}\left(y\right)=x$ gives:
$$x$$

Conclusion
Since both operations yield the same result for any arbitrary element in the domain:
$${(g\circ f)}^{-1}\left(z\right)=(f^{-1}\circ g^{-1})\left(z\right)=x$$
We conclude that:
$${(g\circ f)}^{-1}=f^{-1}\circ g^{-1}$$