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The Proof of Theorem \(6.21 .\) Use the ideas from Exercise (9) to prove Theorem \(6.21 .\) Let \(A, B,\) and \(C\) be nonempty sets and let \(f: A \rightarrow B\) and \(g: B \rightarrow C\) (a) If \(g \circ f: A \rightarrow C\) is an injection, then \(f: A \rightarrow B\) is an injection. (b) If \(g \circ f: A \rightarrow C\) is a surjection, then \(g: B \rightarrow C\) is a surjection.

Short Answer

Expert verified
In summary, we have shown that: (a) If the composition function \(g \circ f: A \rightarrow C\) is an injection, then the function \(f: A \rightarrow B\) is an injection. (b) If the composition function \(g \circ f: A \rightarrow C\) is a surjection, then the function \(g: B \rightarrow C\) is a surjection.

Step by step solution

01

(a) Proof of injection:#

We will prove the statement by contradiction: Let's assume that if g∘f is an injection, but f is not an injection. 1. Recall the definition of an injection: A function is an injection if and only if for all x, y ∈ A, if f(x) = f(y) implies x = y. 2. Since f is not an injection, there must be at least one pair x, y ∈ A (x ≠ y) such that f(x) = f(y). 3. Now consider the composition function g∘f. For this pair (x, y), we have: g(f(x)) = g(f(y)) 4. Since g∘f is an injection, and by definition, it means x = y, which is a contradiction to our initial assumption. Hence, if the composition function g∘f: A → C is an injection, then the function f: A → B is an injection.
02

(b) Proof of surjection:#

We will prove the statement directly: If g∘f is a surjection, then g is a surjection. 1. Recall the definition of a surjection: A function is a surjection if and only if for every element of C, there exists an element of A such that g∘f(x) = c. 2. Let c ∈ C. Since g∘f is a surjection, there exists an element x ∈ A such that: g∘f(x) = g(f(x)) = c 3. Let b = f(x), where b ∈ B. Thus, we have: g(b) = g(f(x)) = c 4. This shows that for every element c ∈ C, there exists an element b ∈ B such that g(b) = c, which implies that g is a surjection. In conclusion, if g∘f: A → C is a surjection, then the function g: B → C is a surjection.

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Key Concepts

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

Injection and Surjection
In mathematics, understanding the concepts of injective and surjective functions is crucial for grasping the nature of different mappings between sets. An **injection** (or one-to-one function) is a function where each element of the target set is mapped from a distinct element in the domain set. This means \(f: A \to B\) is an injection if for all \(x, y \in A\), \(f(x) = f(y)\) implies \(x = y\).

On the other hand, a **surjection** (or onto function) is a function where every element of the target set is mapped by some element of the domain set. Formally, a function \(f: A \to B\) is surjective if for every \(b \in B\), there exists at least one \(a \in A\) such that \(f(a) = b\).

In proving mathematical theorems, understanding when a composition of functions retains these properties helps us validate the nature of newly defined mappings.
Function Composition
Function composition refers to combining two functions to form a new function. If we have two functions, \(f: A \to B\) and \(g: B \to C\), then the composition \(g \circ f: A \to C\) combines them into a single function where the output of \(f\) becomes the input of \(g\).

Mathematically, for any element \(x \in A\), \(g \circ f(x) = g(f(x))\). This operation is essential, especially when dealing with complex systems of functions, as it allows us to execute multiple transformations in a single operation.
  • Function composition is associative, meaning that the order of operations does not change the result, \((h \circ (g \circ f)) = ((h \circ g) \circ f)\).
  • However, function composition is generally not commutative, meaning \(g \circ f eq f \circ g\).
Understanding function composition is key in analyzing the properties of composite functions, such as injectiveness and surjectiveness, as seen in various theorems.
Mathematical Logic
Mathematical logic underpins all of mathematics by providing the framework in which mathematical statements are formulated and analyzed. At its core, it involves concepts such as predicates, propositions, and logical connectives like 'and', 'or', 'not', and 'implies'.
  • Logical equivalences allow us to manipulate and transform logical expressions into equivalent forms.
  • Quantifiers such as 'for all' \( (\forall) \) and 'there exists' \( (\exists) \) are crucial for formulating precise definitions and theorems.
In the context of theorem proving, understanding logical constructs helps in structuring proofs, such as proofs by contradiction. This involves assuming the negation of what we want to prove and showing that this assumption leads to a contradiction, thereby proving the original statement. Logical reasoning is the backbone of forming robust mathematical arguments.
Theorem Proving
The art of theorem proving involves using logical reasoning to establish the truth of mathematical statements. Theorems often involve proving that all conditions lead to a conclusion using a sequence of logical steps.

There are various proof techniques:
  • **Direct Proof**: Straightforwardly shows that premises lead directly to the conclusion.
  • **Proof by Contradiction**: Assumes the opposite of what needs to be proven and demonstrates that this assumption leads to a contradiction.
  • **Proof by Induction**: Typically used for statements involving natural numbers, where the base case is established, and an inductive step shows that if a statement holds for an arbitrary case, it holds for the next one.
Proving that function properties like injectiveness and surjectiveness are maintained when composing functions often requires intricate logical steps. These proofs demand a deep understanding of function behavior within the framework of mathematical logic and precision in reasoning.

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Most popular questions from this chapter

In our definition of the composition of two functions, \(f\) and \(g\), we required that the domain of \(g\) be equal to the codomain of \(f\). However, it is sometimes possible to form the composite function \(g \circ f\) even though \(\operatorname{dom}(g) \neq \operatorname{codom}(f) .\) For example, let $$ f: \mathbb{R} \rightarrow \mathbb{R} $$ be defined by \(\quad f(x)=x^{2}+1,\) and let \(g: \mathbb{R}-\\{0\\} \rightarrow \mathbb{R} \quad\) be defined by \(\quad g(x)=\frac{1}{x}\) (a) Is it possible to determine \((g \circ f)(x)\) for all \(x \in \mathbb{R} ?\) Explain. (b) In general, let \(f: A \rightarrow T\) and \(g: B \rightarrow C\). Find a condition on the domain of \(g\) (other than \(B=T\) ) that results in a meaningful definition of the composite function \(g \circ f: A \rightarrow C\).

For each natural number \(k,\) let \(A_{k}\) be a set, and for each natural number \(n\), let \(f_{n}: A_{n} \rightarrow A_{n+1}\) For example, \(f_{1}: A_{1} \rightarrow A_{2}, f_{2}: A_{2} \rightarrow A_{3}, f_{3}: A_{3} \rightarrow A_{4},\) and so on. Use mathematical induction to prove that for each natural number \(n\) with \(n \geq 2,\) if \(f_{1}, f_{2}, \ldots, f_{n}\) are all bijections, then \(f_{n} \circ f_{n-1} \circ \cdots \circ f_{2} \circ f_{1}\) is a bijection and $$ \left(f_{n} \circ f_{n-1} \circ \cdots \circ f_{2} \circ f_{1}\right)^{-1}=f_{1}^{-1} \circ f_{2}^{-1} \circ \cdots \circ f_{n-1}^{-1} \circ f_{n}^{-1} $$

Constructing an Inverse Function. If \(f: A \rightarrow B\) is a bijection, then we know that its inverse is a function. If we are given a formula for the function \(f,\) it may be desirable to determine a formula for the function \(f^{-1}\). This can sometimes be done, while at other times it is very difficult or even impossible. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f(x)=2 x^{3}-7 .\) A graph of this function would suggest that this function is a bijection. (a) Prove that the function \(f\) is an injection and a surjection. Let \(y \in \mathbb{R}\). One way to prove that \(f\) is a surjection is to set \(y=f(x)\) and solve for \(x\). If this can be done, then we would know that there exists an \(x \in \mathbb{R}\) such that \(f(x)=y .\) For the function \(f,\) we are using \(x\) for the input and \(y\) for the output. By solving for \(x\) in terms of \(y,\) we are attempting to write a formula where \(y\) is the input and \(x\) is the output. This formula represents the inverse function. (b) Solve the equation \(y=2 x^{3}-7\) for \(x\). Use this to write a formula for \(f^{-1}(y),\) where \(f^{-1}: \mathbb{R} \rightarrow \mathbb{R}\) (c) Use the result of Part (13b) to verify that for each \(x \in \mathbb{R}, f^{-1}(f(x))=\) \(x\) and for each \(y \in \mathbb{R}, f\left(f^{-1}(y)\right)=y\) Now let \(\mathbb{R}^{+}=\\{y \in \mathbb{R} \mid y>0\\} .\) Define \(g: \mathbb{R} \rightarrow \mathbb{R}^{+}\) by \(g(x)=e^{2 x-1}\) (d) Set \(y=e^{2 x-1}\) and solve for \(x\) in terms of \(y\). (e) Use your work in Exercise (13d) to define a function \(h: \mathbb{R}^{+} \rightarrow \mathbb{R}\). (f) For each \(x \in \mathbb{R},\) determine \((h \circ g)(x)\) and for each \(y \in \mathbb{R}^{+},\) determine \((g \circ h)(y)\) (g) Use Exercise (6) to explain why \(h=g^{-1}\).

Let \(R_{5}=\\{0,1,2,3,4\\}\) (a) Define \(f: R_{5} \rightarrow R_{5}\) by \(f(x)=x^{2}+4(\) mod 5\()\) for all \(x \in R_{5} .\) Write the inverse of \(f\) as a set of ordered pairs and explain why \(f^{-1}\) is not a function. (b) Define \(g: R_{5} \rightarrow R_{5}\) by \(g(x)=x^{3}+4(\) mod 5\()\) for all \(x \in R_{5} .\) Write the inverse of \(g\) as a set of ordered pairs and explain why \(g^{-1}\) is a function. (c) Is it possible to write a formula for \(g^{-1}(y),\) where \(y \in R_{5} ?\) The answer to this question depends on whether or not is possible to define a cube root of elements of \(R_{5} .\) Recall that for a real number \(x,\) we define the cube root of \(x\) to the real number \(y\) such that \(y^{3}=x .\) That is, $$ y=\sqrt[3]{x} \text { if and only if } y^{3}=x $$ Using this idea, is it possible to define the cube root of each number in \(\mathbb{Z}_{5} ?\) If so, what are \(\sqrt[3]{0}, \sqrt[3]{1}, \sqrt[3]{2}, \sqrt[3]{3},\) and \(\sqrt[3]{4}\). (d) Now answer the question posed at the beginning of Part (c). If possible, determine a formula for \(g^{-1}(y)\) where \(g^{-1}: R_{5} \rightarrow R_{5}\)

(a) Let \(S=\\{1,2,3,4\\} .\) Define \(F: S \rightarrow \mathbb{N}\) by \(F(x)=x^{2}\) for each \(x \in S\). What is the range of the function \(F\) and what is \(F(S) ?\) How do these two sets compare? Now let \(A\) and \(B\) be sets and let \(f: A \rightarrow B\) be an arbitrary function from \(A\) to \(B\). (b) Explain why \(f(A)=\operatorname{range}(f)\). (c) Define a function \(g: A \rightarrow f(A)\) by \(g(x)=f(x)\) for all \(x\) in \(A\). Prove that the function \(g\) is a surjection.

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