Question

Let function \(\mathrm{f}\) be a subset of \(\mathrm{A} \times \mathrm{A}\). Prove that, for every function \(\mathrm{f}_{1}\) and \(\mathrm{f}_{2}\) which are subsets of \(\mathrm{A} \times \mathrm{A}\), if \(\mathrm{f} \circ \mathrm{f}_{1}=\mathrm{f} \circ \mathrm{f}_{2}\) then \(\mathrm{f}_{1}=\mathrm{f}_{2}\) if and only if \(\mathrm{f}\) is injective.

Short answer

In conclusion, we have proven that for every function \(f_1\) and \(f_2\) which are subsets of \(A \times A\), if \(f \circ f_1 = f \circ f_2 \), then \(f_1 = f_2\) if and only if \(f\) is injective by showing the necessary and sufficient conditions. We demonstrated that if \(f\) is injective and \(f \circ f_1 = f \circ f_2\), then \(f_1 = f_2\). Furthermore, if \(f\) is not injective, there exist a pair of functions \(f_1\) and \(f_2\) such that \(f \circ f_1 = f \circ f_2\), but \(f_1 ≠ f_2\).

Step-by-step solution

Statement 1: If f is injective and f ∘ f1 = f ∘ f2, then f1 = f2.

Suppose f is injective and f ∘ f1 = f ∘ f2. To show that f1 = f2, we need to prove that for any x ∈ A, f1(x) = f2(x). Let x ∈ A, and let y1 = f1(x) and y2 = f2(x). Since f ∘ f1 = f ∘ f2, we have that f(y1) = f( f1(x) ) = f( f2(x) ) = f(y2). By definition, if f is injective, then this equality f(y1) = f(y2) implies that y1 = y2. Therefore, f1(x) = f2(x). Since this holds for any x ∈ A, we can conclude that f1 = f2.

Statement 2: If f is not injective, then there exists a pair of functions f1 and f2 such that f ∘ f1 = f ∘ f2, but f1 ≠ f2.

Suppose f is not injective. Then there exist x1, x2 ∈ A such that x1 ≠ x2 but f(x1) = f(x2). Define two functions f1, f2 as follows: f1(x) = x1 for all x ∈ A, and f2(x) = x2 for all x ∈ A. Clearly, f1 ≠ f2 since f1(x) ≠ f2(x) for any x ∈ A. Now let's look at the composition of f with f1 and f2: f ∘ f1(x) = f(f1(x)) = f(x1) and f ∘ f2(x) = f(f2(x)) = f(x2). Since f(x1) = f(x2), we have that f ∘ f1(x) = f ∘ f2(x) for all x ∈ A. Thus, we found a pair of functions f1 and f2 such that f ∘ f1 = f ∘ f2, but f1 ≠ f2. In conclusion, we have shown the necessary implications and the sufficient conditions for the statement to hold. Therefore, if f ∘ f1 = f ∘ f2, then f1 = f2 if and only if f is injective.

Key concepts

Function Composition

In mathematics, function composition is a powerful concept where one function's output becomes another function's input. Think of it like a Rube Goldberg machine, where the end of one step triggers the start of the next. Formally, if you have two functions, say g and h, composing them creates a new function g ∘ h, read as 'g composed with h'. This means for any input x, the resulting output is g(h(x)).

The exercise presented is an investigation into the behavior of function compositions when the starting function, let's denote it as f, is injective. Injectivity is one of the key properties that can affect how a composition behaves, and it's crucial in establishing the link between f₁ and f₂ when composed with f. Injective functions, sometimes also called one-to-one, ensure that no two different inputs produce the same output, which is central to proving the given statement.

Bijection

When we study functions, we often encounter the term bijection, which is a keystone concept in the field of discrete mathematics. A function is bijective if it is both injective (one-to-one) and surjective (onto). In simpler terms, a bijective function creates a perfect 'pairing' between the members of its domain and range—every element in the domain is matched with a unique element in the range, with no duplicates or omissions.

Understanding that a function is bijective can have powerful implications. For instance, it guarantees an inverse: you can perfectly backtrack from the output to the input. While our specific exercise isn't directly concerned with bijections, recognizing when a function is injective—as is necessary for a bijection—helps shed light on the behavior of function compositions.

Function Properties

When investigating functions in discrete mathematics, we must understand their properties to predict how they'll behave under various operations such as composition. The properties like injective, surjective, and bijective play vital roles.

An injective function ensures that different inputs lead to different outputs, implying a kind of uniqueness for each element in the domain. The exercise asks us to use this property to deduce that two functions, which when composed with an injective function yield the same result, must indeed be the same function. This wouldn't necessarily hold if our function wasn't injective because then, the aforementioned uniqueness would be lost.

Discrete Mathematics

Lastly, the broad field in which we study these concepts is called discrete mathematics. It is an area of mathematics focused on the study of objects and ideas that can be distinctly separated and counted. Functions, along with concepts like sets, logic, and probability, are fundamental to discrete mathematics. They offer a language to formalize and analyze the structure and behavior of these countable entities.

In the context of our exercise, we are essentially proving a theorem within the realm of discrete mathematics. By using logical reasoning and the properties of functions, we establish whether a particular condition about function composition holds. It's through such meticulous work that discrete mathematics allows us to build complex ideas out of straightforward, discrete building blocks.