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Parts 1 and 2 , together, tell us that if \(g \circ f\) is bijective, then \(f\) is injective and \(g\) is surjective. Is the converse of this statement true: If \(f\) is injective and \(g\) surjective, is \(g=f\) bijective? (If "yes," prove it; if "no," give a counterexample.)

Short Answer

Expert verified
No, if \( f \) is injective and \( g \) is surjective, \( g \circ f \) may not be bijective.

Step by step solution

01

Understand the Definitions

To tackle the problem, first recall the definitions: - A function is **injective** (or one-to-one) if different inputs produce different outputs. - A function is **surjective** (or onto) if every element in the target set is the image of at least one element from the domain. - A function is **bijective** if it is both injective and surjective.
02

Analyze the Property of Composition

The problem states that if the composition \( g \circ f \) is bijective, \( f \) is injective and \( g \) is surjective. We need to explore if the converse is true: if \( f \) is injective and \( g \) is surjective, is \( g \circ f \) bijective?
03

Formulate Possible Scenarios

Since \( f \) is injective, for every distinct input \( x_1 \) and \( x_2 \), \( f(x_1) eq f(x_2) \). If \( g \) is surjective, every element in the codomain of \( g \) has a pre-image under \( g \), meaning every value of \( g(y) \) corresponds to at least one \( f(x) \).
04

Consider a Counterexample

Consider the functions \( f: \{1, 2\} \rightarrow \{a, b\} \) defined by \( f(x) = a\) for both \( x = 1 \) and \( x = 2 \), making \( f \) not injective, but for this step, suppose \( f \) is injective. Now, let \( g: \{a, b\} \rightarrow \{c\} \) be defined by \( g(a) = g(b) = c \). \( g \) is not injective, but it is surjective to set \( \{c\} \). The composition \( g \circ f \) maps both 1 and 2 to \( c \).
05

Conclusion from Counterexample

The counterexample illustrates that \( g \circ f \) is not injective as multiple inputs map to the same output \( c \). Hence, even though \( f \) can be injective and \( g \) surjective individually (ignoring our setup where \( f \) is not), their composition \( g \circ f \) is not necessarily bijective.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Function
An injective function, also commonly known as a one-to-one function, has a unique characteristic: each input is mapped to a distinct output. In simpler terms, no two different inputs will have the same output in an injective function. Imagine you have a perfect lock and key system; each key has its own lock. When you try to open a lock, only one specific key will work. This is akin to how an injective function operates. If we have a function \( f: A \rightarrow B \), it is injective if for any two elements \( x_1 \) and \( x_2 \) in set \( A \), whenever \( f(x_1) = f(x_2) \) it must be that \( x_1 = x_2 \).

For example, the function \( f(x) = 2x \) from real numbers to real numbers is injective because if \( 2x_1 = 2x_2 \), then \( x_1 = x_2 \) must hold true. Understanding injective functions helps in analyzing the composition of two functions, especially when determining whether a composed function is bijective.
Surjective Function
A surjective function, known as an onto function, ensures that every possible output is covered by at least one input from its domain. This means, for a function \( g: B \rightarrow C \), each element of set \( C \) should have a pre-image in set \( B \). Think of surjective functions as ensuring a complete task where every employee has at least one task assigned, with no tasks left out.

In mathematical terms, a function \( g \) is surjective if for every element \( c \) in set \( C \), there exists at least one element \( b \) in set \( B \) such that \( g(b) = c \). A simple example is the function \( g(y) = y^2 \), mapping non-negative real numbers onto non-negative real numbers. Every non-negative number in the codomain is the square of some non-negative number.

Understanding surjective functions plays a crucial role in certain scenarios, such as when analyzing the conditions under which a function composition like \( g \circ f \) is bijective.
Bijective Function
A bijective function exhibits the best of both worlds: it is both injective and surjective. This means that every input has a unique output, and every potential output is covered. In essence, a bijective function is like a perfect matchmaking service, where each person is matched with exactly one partner, and no one is left out.

Formally, for a function \( h: A \rightarrow C \) to be bijective, it must be true that for every element \( c \) in set \( C \), there is exactly one element \( a \) in set \( A \) such that \( h(a) = c \). Consequently, a bijective function can be uniquely "reversed" or inverted, meaning you can switch inputs and outputs without loss of information or ambiguity.

Consider the function \( h(x) = x + 3 \) from real numbers to real numbers. This function is bijective because each input corresponds to a unique output (injective), and every real number can be written as \( x + 3 \) (surjective). The properties of bijections are especially significant when discussing function compositions, as a composition like \( g \circ f \) will only be bijective if one function is injective and the other is surjective.

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