Question

Let \(X\) be a set, and let \(\mathcal{F}_{X}\) be the set of all \(I-I\) functions from \(X\) onto \(X\). We have two operations on functions in \(\mathcal{F}_{X}: \circ\) and -1 . Prove the following statements called group axioms. (If the results are already proved in the book, note where to find the proofs.) (a) For all \(F, G \in \mathcal{F}_{X}, F \circ G \in \mathcal{F}\). (b) For all \(F, G, H \in \mathcal{F}_{X},(F \circ G) \circ H=F \circ(G \circ H)\) (Associative Law). (c) For all \(F \in \mathcal{F}_{X}, F \circ I d_{X}=I d_{X} \circ F=F\). (Identity Axiom). (d) For all \(F \in \mathcal{F}_{X}\), there exists an \(F^{-1}\) such that \(F \circ F^{-1}=F^{-1} \circ F=I d_{X}\) (Inverse Axiom).

Short answer

The set \(\mathcal{F}_X\) with composition forms a group, satisfying closure, associativity, identity, and inverses.

Step-by-step solution

Closure Property

To show closure under function composition, consider that for all functions \(F, G \in \mathcal{F}_X\), the composition \(F \circ G\) is defined as \((F \circ G)(x) = F(G(x))\) for all \(x \in X\). Because \(G\) is onto, for every \(y \in X\), there exists \(x \in X\) such that \(G(x) = y\). Since \(F\) is onto, \(F(G(x))=F(y)\) covers all possible outputs of \(X\). Hence, \(F \circ G\) is onto, thus \(F \circ G \in \mathcal{F}_X\).

Associative Law

To prove associativity, take any three functions \(F, G, H \in \mathcal{F}_X\). We want to show that \((F \circ G) \circ H = F \circ (G \circ H)\). For every \(x \in X\), \(((F \circ G) \circ H)(x) = (F \circ G)(H(x)) = F(G(H(x)))\) and \((F \circ (G \circ H))(x) = F((G \circ H)(x)) = F(G(H(x)))\). Both sides are equal, thus the associative law holds.

Identity Element

To demonstrate the existence of an identity element, let \(I d_X\) be the identity function defined by \(I d_X(x) = x\) for all \(x \in X\). For any \(F \in \mathcal{F}_X\), \((F \circ I d_X)(x) = F(I d_X(x)) = F(x)\) and \((I d_X \circ F)(x) = I d_X(F(x)) = F(x)\). Hence, \(I d_X \circ F = F \circ I d_X = F\), showing that \(I d_X\) is indeed the identity element.

Inverse Element

To prove the existence of an inverse for each \(F \in \mathcal{F}_X\), let \(F^{-1}\) be the inverse of \(F\), such that \(F(F^{-1}(x)) = x\) and \(F^{-1}(F(x)) = x\) for all \(x \in X\). Since \(F\) is a bijection (because it's onto and one-to-one), \(F^{-1}\) is well-defined. Therefore, \(F \circ F^{-1} = F^{-1} \circ F = I d_X\), proving that each function \(F\) in \(\mathcal{F}_X\) has an inverse in \(\mathcal{F}_X\).

Key concepts

Bijection

A bijection is a special type of function that perfectly pairs every element in the set with a unique element in another set (or possibly the same set). In the context of this exercise, we're talking about functions from a set \(X\) onto itself, which means every element in \(X\) is both the domain and codomain of the functions.

• One-to-One: For a function to be one-to-one (injective), every element of the domain must map to a different element in the codomain.
• Onto: A function is onto (surjective) if every element in the codomain is mapped to by at least one element in the domain.

Bijections are crucial in understanding group operations because they ensure reversibility and completeness, meaning each element can be uniquely identified and paired.

Function Composition

Function composition is an operation where you combine two functions to form a new one. In our set \(\mathcal{F}_X\), if you have two functions \(F\) and \(G\), composing them as \(F \circ G\) means performing \(G\) first, and then applying \(F\) to the result of \(G\).
Function composition is essential for understanding how different transformations (functions) can be combined to form a single transformation.

• Closure in Composition: If \(F\) and \(G\) are functions from \(X\) to \(X\), their composition \(F \circ G\) is also a function from \(X\) onto \(X\).
• Associative Property: Given functions \(F\), \(G\), and \(H\), it holds that \((F \circ G) \circ H = F \circ (G \circ H)\), meaning the order of composition does not change the outcome.

Inverse Function

An inverse function essentially "undoes" what the original function does. For a function \(F\) to have an inverse, \(F\) must be a bijection. This means \(F\) is both injective and surjective, exactly meeting each element of its codomain once.
The inverse function, denoted as \(F^{-1}\), swaps the roles of inputs and outputs of \(F\). In other words, if \(F(x) = y\), then \(F^{-1}(y) = x\).

• Existence: The existence of this inverse is guaranteed as long as \(F\) is a bijection.
• Inverse Law: For any function \(F\) and its inverse \(F^{-1}\), we have \(F \circ F^{-1} = F^{-1} \circ F = Id_X\), where \(Id_X\) is the identity function.

Identity Function

The identity function is a very straightforward concept. For a set \(X\), the identity function, denoted \(Id_X\), simply returns the same input, \(Id_X(x) = x\) for every \(x\) in \(X\).
In the context of function operations, the identity function acts like the number zero does for addition or one for multiplication. It doesn't change the value of the element it operates on.

• Identity Element: When composing \(Id_X\) with any function \(F\), the original output is maintained, meaning \(Id_X \circ F = F \circ Id_X = F\).
• Role in Groups: For function groups, the identity function ensures no change, maintaining the structure of our system as a neutral element.