Question

Let \(\Phi\) be a mapping from set \(\mathrm{A}\) to set \(\mathrm{B}\). Show: (A) If there exists another map \(\Upsilon\) from \(B\) to \(A\) such that \(\Upsilon \circ \Phi=\mathrm{I}_{\mathrm{A}}\) (where \(\mathrm{I}_{\mathrm{A}}\) is the identity map from \(\mathrm{A}\) to \(\mathrm{A}\) ) then \(\Phi: \mathrm{A} \rightarrow \mathrm{B}\) is an injective map. (B) If there is a mapping \(\theta\) from \(B\) to \(A\) such that \(\Phi \circ \theta=I_{B}\) then \(\Phi: \mathrm{A} \rightarrow \mathrm{B}\) is a surjective map.

Short answer

A mapping \(\Phi: A \to B\) is injective if \(\exists \Upsilon: B \to A\) such that \(\Upsilon \circ \Phi = I_A\), and surjective if \(\exists \theta: B \to A\) such that \(\Phi \circ \theta = I_B\). Proof: 1. Injective: Given \(\Phi(a_1) = \Phi(a_2)\), applying \(\Upsilon\) to both sides and using \(\Upsilon \circ \Phi = I_A\) yields \(a_1 = a_2\). 2. Surjective: Given any \(b \in B\), we have \(a = \theta(b) \in A\) such that \(\Phi(a) = b\), thus \(\Phi\) is surjective.

Step-by-step solution

Understand the given mappings

We have a mapping \(\Phi: A \to B\), and two other mappings, \(\Upsilon: B \to A\) and \(\theta: B \to A\), which fulfill certain conditions with respect to \(\Phi\) and the identity maps on sets A and B.

Recall Identity map and injectivity concepts

The identity map, denoted as \(I_A\) on set A and \(I_B\) on set B, maps each element of the set to itself. A function/mapping is considered injective (also known as one-to-one) if every element of the domain (set A) is mapped to a unique element in the codomain (set B). In other words, for an injective mapping \(\Phi\), \(\Phi(a_1) = \Phi(a_2)\) implies that \(a_1 = a_2\).

Prove injective mapping using the given conditions

We are given that \(\Upsilon \circ \Phi = I_A\), meaning that for all \(a \in A\), \((\Upsilon \circ \Phi)(a) = a\). Now, let there be two elements \(a_1\) and \(a_2\) in A such that \(\Phi(a_1) = \Phi(a_2)\). We need to show that \(a_1 = a_2\) to prove that \(\Phi\) is injective. Apply the function \(\Upsilon\) to both sides of the equation: \(\Upsilon(\Phi(a_1)) = \Upsilon(\Phi(a_2))\) The given condition, \(\Upsilon \circ \Phi = I_A\), implies: \((\Upsilon \circ \Phi)(a_1) = (\Upsilon \circ \Phi)(a_2)\) Hence, \(a_1 = a_2\), which proves that \(\Phi\) is injective.

Recall Surjectivity concept

A function/mapping is considered surjective (also known as onto) if for every element of the codomain (set B), there exists an element in the domain (set A) such that the mapping \(\Phi\) maps that element to the element in the codomain. In other words, for a surjective mapping \(\Phi\), \(\forall b \in B, \exists a \in A\) such that \(\Phi(a) = b\).

Prove surjective mapping using the given conditions

We are given that \(\Phi \circ \theta = I_B\), meaning that for all \(b \in B\), \((\Phi \circ \theta)(b) = b\). Now, let there be an element \(b\) in set B, we need to show the existence of an element \(a\) in set A such that \(\Phi(a) = b\) to prove that \(\Phi\) is surjective. Since \(\Phi \circ \theta = I_B\), so for every \(b \in B\), there exists an element \(a = \theta(b) \in A\) such that: \(\Phi(\theta(b)) = (\Phi \circ \theta)(b) = b\) Hence, we have a corresponding element \(a\) in set A for the given \(b\) such that \(\Phi(a) = b\). This proves that \(\Phi\) is surjective.

Key concepts

Injective Map

When we talk about an injective map, often referred to as a 'one-to-one' function, we are focusing on uniqueness. Injectivity is a quality of a function that ensures no two different elements in the domain map to the same element in the codomain. To put it plainly, every peg fits into its own unique hole and no two pegs share the same hole.

Let’s consider the symbols \(a_1\) and \(a_2\) as two distinct elements from the domain set A. If our mapping \(\Phi\) is injective, and if \(\Phi(a_1) = \Phi(a_2)\), we can confidently conclude that \(a_1 = a_2\). This exclusive pairing is much like having a unique key for every lock; no two keys can open the same lock. In the context of the exercise, the existence of the mapping \(\Upsilon\) that when composed with \(\Phi\) yields the identity map \(\mathrm{I}_{\mathrm{A}}\) is what gives us the proof that \(\Phi\) is indeed injective.

Surjective Map

Moving on to the concept of a surjective map, which is sometimes called an 'onto' function, we shift our focus from uniqueness to completeness. A surjective function is one where every element in the codomain has a pre-image in the domain. Imagine that every person at a party must get a slice of cake. If the mapping from slices of cake (domain) to partygoers (codomain) is surjective, then everyone gets a slice, no one is left out, and all the cake is distributed.

In our exercise, we're given that \(\Phi \circ \theta = I_{B}\), which intuitively means that everyone in B gets attention from A through \(\Phi\). Specifically, for every \(b \) in B, there must be an \(a \) in A such that \(\Phi(a) = b\). So, just as every partygoer gets their piece of cake, each element in B finds its partner in A, establishing the surjective nature of the mapping \(\Phi\).

Identity Map

The identity map, often symbolized as \(I\), is like a mirror in the world of functions. It reflects every element back to itself without any change. In other words, \(I_A\) on set A means that every \(a \) in A satisfies \(I_A(a) = a\), and every \(b \) in B satisfies \(I_B(b) = b\) when considering \(I_B\).

This concept may seem trivial, but it's an essential piece in understanding more complex function compositions in mathematics. It is like having an exact clone of every individual in a community; each person is paired with themselves, and there is a one-to-one correspondence, reflecting the essence of an identity map.

Composition of Functions

Picture a relay race where a baton is passed from one runner to the next. The composition of functions operates similarly, but with inputs and outputs. If we have two functions, say \(f\) and \(g\), their composition, noted as \(g \circ f\), means that we first apply \(f\) to an element and then apply \(g\) to the result of \(f\).

Consider the act of painting a sculpted pot. The sculptor shapes it (function \(f\)), and then the painter adds color (function \(g\)). The composition \(g \circ f\) transforms clay into a painted pot in two steps. In the context of our exercise, the composition of functions allows us to investigate deeper properties of the map \(\Phi\), leading to the understanding of its injective and surjective nature.