Problem 1 - A Book of Abstract Algebra

Question

Prove that if \(g \circ f\) is injective, then \(f\) is injective.

Step-by-Step Solution

Answer

If \(g \circ f\) is injective, then \(f\) is injective because distinct inputs must map to distinct outputs.

Step 1: Understanding Injective Functions

A function is said to be injective (or one-to-one) if it maps distinct elements of its domain to distinct elements of its codomain. This means that for a function \(f: A \to B\), if \(f(a_1) = f(a_2)\) implies \(a_1 = a_2\) for all \(a_1, a_2 \in A\), then \(f\) is injective.

Step 2: Define the Composition Function

Let \(g: B \to C\) and \(f: A \to B\) be functions. The composition \(g \circ f\) is a function from \(A\) to \(C\) defined as \((g \circ f)(x) = g(f(x))\). This means for each \(x \in A\), \(g \circ f\) maps \(x\) to \(g(f(x))\).

Step 3: Assume \(g \circ f\) is Injective

According to the problem, \(g \circ f\) is injective. This means for any \(x_1, x_2 \in A\), if \((g \circ f)(x_1) = (g \circ f)(x_2)\), then \(x_1 = x_2\). More clearly, if \(g(f(x_1)) = g(f(x_2))\), then \(x_1 = x_2\).

Step 4: Prove \(f\) is Injective Using Given Condition

Suppose for \(f\), there exist elements \(a_1, a_2 \in A\) such that \(f(a_1) = f(a_2)\). Then, \(g(f(a_1)) = g(f(a_2))\). Since \(g \circ f\) is injective, this implies \(a_1 = a_2\). Hence, when \(f(a_1) = f(a_2)\), we must have \(a_1 = a_2\). This proves that \(f\) is injective.

Key Concepts

Composition of Functions

When we talk about composition of functions, what we really mean is feeding one function into another. Imagine two functions: \(f\) and \(g\). The composition is written as \(g \circ f\). This denotes a function where you first apply \(f\) to an input, and then feed the result into \(g\).
Here's a simple example to illustrate:

• Suppose \(f(x) = 2x\) and \(g(y) = y + 1\).
• For \(g \circ f\), our input \(x\) first goes into \(f\), doubling it to \(2x\).
• Then, \(2x\) becomes the input for \(g\), which adds one, resulting in \(2x + 1\).

The composition process helps us build more complex functions from simpler ones, making it a powerful tool in mathematical analysis and applications.

One-to-One Function

An injective function, often referred to as a one-to-one function, has a unique way it handles its mapping. It assigns each element in its domain a unique result in its codomain. A key property of injective functions is that if two elements in the domain map to the same element in the codomain, those two elements must be identical.
In simpler terms:

• If \(f(x_1) = f(x_2)\), then it must be true that \(x_1 = x_2\).
• This behavior prevents different inputs from "colliding" into the same output.

Think of injective functions like a perfectly functioning vending machine where each button gives a different snack. Pressing the same button will always give the same snack, ensuring no mix-ups!

Function Mapping

Function mapping is how we describe the relation between two sets, the domain and the codomain. When a function \(f: A \to B\) maps set \(A\) to set \(B\), for each element in \(A\), there is a corresponding element in \(B\). This creates a connection, or mapping, between inputs and outputs.
To visualize function mapping:

• Picture arrows leading from each element in the set \(A\) to some element in \(B\).
• If the arrows never point to the same element twice, your function is injective.

This systematic approach helps in understanding how inputs relate to outputs, and dictates the characteristics of different functions. Mapping explains the behavior of functions, whether they are injective, surjective, or bijective.

Injective Proof

An injective proof demonstrates that a function is one-to-one. The exercise showcases a clever proof involving function composition. You've already learned that if \(g \circ f\) is injective, then clearly, \(f\) must also be injective. Here is why:

• Start by assuming \(g(f(x_1)) = g(f(x_2))\).
• If \(g \circ f\) is injective, then \(x_1 = x_2\).
• Suppose \(f(x_1) = f(x_2)\) since \(g(f(x_1)) = g(f(x_2))\).
• According to the injectivity of \(g \circ f\), it confirms \(x_1 = x_2\).

Thus, since different inputs could not lead to the same output in \(g \circ f\), \(f\) must be injective, demonstrating \(f\)'s consistency as a mapping function.