Composite Function of One-One Functions

The question asks about the nature of the composite function \(g \circ f\) when both \(f\) and \(g\) are one-one functions.

Let's first understand what a one-one function means.

A function \(h : X \to Y\) is called one-one (or injective) if distinct elements in the domain \(X\) are mapped to distinct elements in the codomain \(Y\). Mathematically, this means that if \(h(x_1) = h(x_2)\) for any \(x_1, x_2 \in X\), then it must imply that \(x_1 = x_2\).

We are given two functions:

• \(f : A \to B\), and \(f\) is one-one.
• \(g : B \to C\), and \(g\) is one-one.

The composite function is \(g \circ f : A \to C\), defined by \((g \circ f)(x) = g(f(x))\) for all \(x \in A\).

Proving \(g \circ f\) is One-One

To prove that the composite function \(g \circ f\) is one-one, we need to show that if we assume \((g \circ f)(x_1) = (g \circ f)(x_2)\) for \(x_1, x_2 \in A\), it must lead to the conclusion \(x_1 = x_2\).

Let's start with the assumption:

Assume \((g \circ f)(x_1) = (g \circ f)(x_2)\) for some \(x_1, x_2 \in A\).

By the definition of the composite function, this means:

\(g(f(x_1)) = g(f(x_2))\)

Let \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\). Both \(y_1\) and \(y_2\) are elements in the set \(B\). The equation now looks like:

\(g(y_1) = g(y_2)\)

Since we are given that the function \(g : B \to C\) is one-one, the property of one-one functions tells us that if \(g(y_1) = g(y_2)\), then it must be that \(y_1 = y_2\).

Substituting back the expressions for \(y_1\) and \(y_2\), we get:

\(f(x_1) = f(x_2)\)

Now, we use the fact that the function \(f : A \to B\) is also one-one. The property of one-one functions applied to \(f\) tells us that if \(f(x_1) = f(x_2)\), then it must be that \(x_1 = x_2\).

So, starting with \((g \circ f)(x_1) = (g \circ f)(x_2)\), we have successfully concluded that \(x_1 = x_2\). This matches the definition of a one-one function.

Therefore, the composite function \(g \circ f : A \to C\) is one-one.

Summary

If both \(f\) and \(g\) are one-one functions, their composition \(g \circ f\) is also a one-one function. This proof relies directly on the definition of one-one functions applied sequentially to \(g\) and then to \(f\).

The options provided were:

1. One-one and onto both
2. One-one
3. Neither One-One, nor onto
4. Onto

Based on our proof, the composite function \(g \circ f\) is indeed one-one. Whether it is onto depends on the nature of the functions and sets, which is not given in the problem. However, the problem specifically asks about the characteristic that is guaranteed by both \(f\) and \(g\) being one-one, which is that \(g \circ f\) is one-one.