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Chapter 13: Problem 6

Sind \(g: X \rightarrow Y\) und \(f: Y \rightarrow Z\) gegeben, so ist \((f \circ g)^{-1}(C)=g^{-1}\left(f^{-1}(C)\right)\) f眉r jedes \(C \subset Z\). Sind 眉berdies \(\mathrm{g}\) und \(f\) bijektiv, so ist auch \(f \circ \mathrm{g}\) bijektiv und \((f \circ \mathrm{g})^{-1}=\mathrm{g}^{-1} \circ f^{-1}\)

Short Answer

Expert verified
The statement \((f \circ g)^{-1}(C) = g^{-1}(f^{-1}(C))\) holds, and if \(g\) and \(f\) are bijective, \(f \circ g\) is bijective with \((f \circ g)^{-1} = g^{-1} \circ f^{-1}\).

Step by step solution

01

Understanding the Composition of Functions

Given two functions, \( g: X \rightarrow Y \) and \( f: Y \rightarrow Z \), the composition of these functions, \( f \circ g \), maps \( X \) directly to \( Z \). The statement we need to show is \( (f \circ g)^{-1}(C) = g^{-1}(f^{-1}(C)) \) for every subset \( C \subset Z \). This means the preimage via the composition is the same as the preimage via each function separately.
02

Breaking Down the Preimage Statement

For any subset \( C \subset Z \), the set \( (f \circ g)^{-1}(C) \) consists of all elements in \( X \) that are mapped in the image of \( C \) by \( f \circ g \). We need to prove this is equivalent to \( g^{-1}(f^{-1}(C)) \), which means taking elements in \( Z \), finding their preimage under \( f \), and then finding the preimage of that result under \( g \).
03

Proving \( (f \circ g)^{-1}(C) = g^{-1}(f^{-1}(C)) \)

An element \( x \in X \) is in \( (f \circ g)^{-1}(C) \) iff \( (f \circ g)(x) \in C \). Since \( (f \circ g)(x) = f(g(x)) \), \( x \) must satisfy \( g(x) \in f^{-1}(C) \). Thus, \( x \in g^{-1}(f^{-1}(C)) \). Similarly, if \( x \in g^{-1}(f^{-1}(C)) \), then \( g(x) \in f^{-1}(C) \), so \( f(g(x)) \in C \). Therefore, \( x \in (f \circ g)^{-1}(C) \). This confirms both sides are equal.
04

Proving Bijectivity of Composition

Given that \( g \) and \( f \) are bijective functions, they have inverses \( g^{-1} \) and \( f^{-1} \). The composition of bijective functions, \( f \circ g \), is also bijective. A bijective function has both injective and surjective properties, establishing a one-to-one correspondence between the domains and codomains.
05

Establishing the Inverse of the Composite Function

We need to show \( (f \circ g)^{-1} = g^{-1} \circ f^{-1} \). For any element \( z \in Z \), \( x = (f \circ g)^{-1}(z) \) implies \( (f \circ g)(x) = z \). Due to the bijectivity, \( g(g^{-1}(f^{-1}(z))) = f^{-1}(z) \) and \( f(f^{-1}(z)) = z \). Hence, \( x = g^{-1}(f^{-1}(z)) \), confirming the inverse formula.

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Key Concepts

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

Bijective Function
A bijective function is one that is both injective and surjective, meaning every element in the codomain is mapped by exactly one element in the domain. This allows bijective functions to have an inverse function because there is a perfect one-to-one correspondence between both sets involved.

The injective component, also called "one-to-one," ensures that no two different elements in the domain map to the same element in the codomain. Alternatively, the surjective component, or "onto," guarantees that every element in the codomain has a preimage in the domain.

Here are some key points to remember about bijective functions:
  • Every bijective function can be reversed, meaning there exists an inverse function.
  • Bijects the domain perfectly onto the entire codomain.
  • The composition of two bijective functions is also bijective.
Preimage in Set Theory
The preimage concept in set theory revolves around finding all elements in the domain which map to a particular subset in the codomain by a given function. Simply put, if you know the output, you're finding all possible inputs that lead to that output.

In our problem, we are considering the composition of two functions, and how to trace a subset from the codomain of the resulting function back to the original domain of the first function:
  • If function composition maps directly from one domain to another, the preimage helps us find how that mapping was achieved.
  • To apply the definitions in practice, one uses the preimages of the inner function first and then processes these with the outer function.
This step-by-step approach ensures clarity and accuracy in set mapping operations, addressing each function layer individually.
Inverse Function
An inverse function reverses the actions of a given function, effectively swapping each output back to its original input. For an inverse function to exist, the original function must be bijective, which ensures that each input corresponds to one unique output and vice-versa.

Finding the inverse of a function involves switching the roles of inputs and outputs:
  • For a function \( f: A \to B \), its inverse function \( f^{-1}: B \to A \) maps each element of \( B \) back to its original element in \( A \).
  • The notation \( f^{-1} \) denotes the inverse; this should not be confused with reciprocal, especially in non-numeric contexts.
In our exercise, showing that \( (f \circ g)^{-1} = g^{-1} \circ f^{-1} \) required demonstrating that this swapped process works equivalently for the composite function.
Mathematical Proof
Mathematical proofs provide the logical foundation and verification for statements or formulas. They are crucial for establishing the truth in mathematical reasoning, ensuring that any conclusions drawn are accurate and reliable.

In the exercise, proof was necessary for two key assertions:
  • Establishing that preimages work correctly through consecutive functions.
  • Demonstrating that the composition of bijective functions remains bijective and how this affects their inverses.
Breaking down the equations and testing logical equivalencies helped in proving the given mathematical statements by showing each step as a natural consequence of the previous. Mathematical proofs are iterative, often requiring incremental validation of each logical step until the final result holds as a conclusion.

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Most popular questions from this chapter

Zeige an Selbstabbildungen von \(X\), daB \(f \circ g\) nicht mit \(g \circ f\) 眉bereinzustimmen braucht: F眉r die Komposition gilt kein Kommutativgesetz. Dagegen gilt das Assoziativgesetz \(f \circ(g \circ h)=(f \circ g) \circ h\), falls das linksstehende Kompositum existiert.

Gegeben sei die Funktion \(f: X \rightarrow Y . A\) bzw \(B\) stehe f眉t Teilmengen aus \(X\) bzw \(Y\), ferner sei \(A^{\prime}:=X \backslash A\) und \(B^{\prime}:=Y \backslash B\) Zeige: a) \(A_{1} \subset A_{2} \rightarrow f\left(A_{1}\right) \subset f\left(A_{2}\right), B_{1} \subset B_{2} \rightarrow f^{-1}\left(B_{1}\right) \subset f^{-1}\left(B_{2}\right) ;\) d) \(f^{-1}\left(B^{\prime}\right)=\left(f^{-1}(B)\right)^{\prime}\)

Sei \(f: X \rightarrow Y\) gegeben. Zeige: a) \(f\) ist injektiv \(\leftrightarrow\) es gibt ein \(g: Y \rightarrow X\) mit \(g \circ f=i d_{x}\) b) \(f\) ist surjektiv \(\leftrightarrow\) es gibt ein \(h: Y \rightarrow X\) mit \(f \circ h=i d_{\gamma}\) c) \(f\) ist bijektiv \(\Leftrightarrow\) es gibt ein \(g: Y \rightarrow X\) mit \(g \circ f=i d_{x}, f \circ g=i d_{Y}\) Dieses g ist, falls vorhanden, eindeutig bestimmt
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