Animated Solution for Mathematics - Functions: Consider function $f:A\to B$ and $g:B\to C(A,B,C\subseteq \mathbf{R})$ such that $(g\circ f{)}^{-1}$ exists then:

Unlocking the Composite Function

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of how functions communicate.

When we talk about a composite function $(g\circ f):A\to C$, we are looking at a two-stage journey. An element $a\in A$ travels through $f$ to become $b\in B$, and then travels through $g$ to arrive at $c\in C$.

When we are told that $(g\circ f{)}^{-1}$ exists, we are being given a profound piece of information: this entire journey is perfectly reversible.

Before we dive into the mechanics, let us ground ourselves in the definition of an inverse. For any function to possess an inverse, it must be bijective.

This means it must be both one-one (injective) and onto (surjective). If a function is not one-one, it 'collapses' information, making it impossible to trace back the path.

If it is not onto, it leaves parts of the codomain 'unreachable,' making the inverse function incomplete. Therefore, because $(g\circ f)$ is invertible, we know with absolute certainty that $(g\circ f)$ is both one-one and onto. This is our bedrock.

Protecting the Uniqueness

Let us examine the first stage: $f:A\to B$. Could $f$ be many-one? Imagine for a moment that $f$ is not one-one.

This would mean there exist two distinct elements $x_1,x_2\in A$ such that $f(x_1)=f(x_2)=y$. If we then pass these through $g$, we get:

Suddenly, the composite function $(g\circ f)$ maps two different inputs to the same output. But wait! We already established that $(g\circ f)$ is one-one.

This is a direct contradiction. Therefore, the only way to preserve the integrity of the composite function is for $f$ to be one-one. Every distinct input in $A$ must map to a distinct output in $B$. There is no room for 'collisions' in the first stage.

Ensuring Coverage

Now, let us turn our attention to the second stage: $g:B\to C$. We know that $(g\circ f)$ is onto, which means for every element $c\in C$, there exists at least one $a\in A$ such that:

Let us break this down. We can write this as $g(f(a))=c$. If we define $f(a)=b$, where $b$ is an element in $B$, then the equation becomes:

This tells us that for every single element $c$ in the final set $C$, we have found a corresponding element $b$ in the intermediate set $B$ that maps to it via $g$. This is the very definition of an onto function. Thus, $g$ must be onto. It must cover the entire set $C$.

By analyzing the constraints imposed by the existence of the inverse, we have successfully dissected the roles of $f$ and $g$. We found that $f$ acts as the guardian of uniqueness, ensuring that no two inputs are confused, while $g$ acts as the guardian of coverage, ensuring that no part of the final destination is left behind.

We have proven that $f$ must be one-one and $g$ must be onto. This is the elegant symmetry of composite functions.

As you move forward in your JEE preparation, remember this: whenever you see an inverse, look for the bijectivity. It is the key that unlocks the structure of the entire system. Keep questioning, keep visualizing, and most importantly, keep enjoying the beautiful logic of mathematics.