Let F Be A Function From A To B. (a) Show That If F Is Injective And EA, Then F 1(f(E))=E. Give An Example

Let F Be A Function From A To B. (a) Show That If F Is Injective And EA, Then F⁻¹(f(E))=E. Give An Example

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Introduction

Mathematics is a language of relationships, structures, and functions that connect various sets and elements within them. Understanding the properties of functions, particularly injectivity and surjectivity, is crucial in fields ranging from algebra to analysis.

This article explores a specific property involving functions between sets, focusing on the relationship between a function \(F : A \to B\), its inverse image, and the concepts of injectivity and the image of a subset. We will demonstrate that if \(F\) is injective and onto the image of a subset \(E \subseteq A\), then the inverse image of the image of \(E\) under \(F\) equals \(E\). Additionally, an illustrative example will clarify this concept.

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Understanding the Fundamental Concepts

Before delving into the proof and example, let's clarify the key definitions involved:

1. Functions and Sets

• Function \(F : A \to B\): A rule that assigns each element of set \(A\) to exactly one element of set \(B\).
• Subset \(E \subseteq A\): A collection of elements within \(A\).
• Image of \(E\) under \(F\): Denoted \(F(E)\), it is the set \(\{F(e) : e \in E\}\).

2. The Inverse Image (Pre-image) \(F^{-1}\)

• For a subset \(Y \subseteq B\), the inverse image \(F^{-1}(Y)\) is defined as:

\[ F^{-1}(Y) = \{a \in A : F(a) \in Y\} \]

• Notably, \(F^{-1}(F(E))\) contains all elements in \(A\) that map into \(F(E)\).

3. Injective (One-to-One) Functions

• Injectivity: A function \(F\) is injective if:

\[ \text{For all } a1, a2 \in A, \quad \text{if } F(a1) = F(a2), \text{ then } a1 = a2 \]

• Intuition: No two different elements in \(A\) map to the same element in \(B\).

4. Surjective (Onto) and The Notation \(EA\)

• Surjectivity: A function \(F : A \to B\) is surjective if:

\[ \text{For every } b \in B, \quad \text{there exists } a \in A \text{ such that } F(a) = b \]

• EA notation: In some contexts, \(EA\) may denote that \(F\) is onto its image, or that the image of \(E\) under \(F\) is the entire set \(F(E)\). For clarity, in this context, it indicates the mapping is onto the subset \(F(E)\).

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Statement of the Theorem

Theorem:
Let \(F : A \to B\) be a function. If \(F\) is injective and the restriction of \(F\) to \(E \subseteq A\) is onto its image \(F(E)\) (i.e., \(F|_E : E \to F(E)\) is surjective), then:

\[
F^{-1}(F(E)) = E
\]

Furthermore, we will illustrate this with a concrete example.

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Proof of the Theorem

To establish the equality \(F^{-1}(F(E)) = E\), we need to demonstrate two inclusions:

1. \(E \subseteq F^{-1}(F(E))\)

• Argument:
• Take any element \(e \in E\).
• Since \(F(e) \in F(E)\) by definition, it follows that:

\[ e \in F^{-1}(F(E)) \]

• Therefore:

\[ E \subseteq F^{-1}(F(E)) \]

2. \(F^{-1}(F(E)) \subseteq E\)

• Assumption: \(F\) is injective and \(F|_E\) is onto \(F(E)\).
• Suppose \(a \in F^{-1}(F(E))\). Then:

\[ F(a) \in F(E) \]

• So, there exists some \(e \in E\) such that:

\[ F(a) = F(e) \]

• Since \(F\) is injective, the equality \(F(a) = F(e)\) implies:

\[ a = e \]

• Conclusion:

\[ a \in E \]

• Hence:

\[ F^{-1}(F(E)) \subseteq E \]

Combining both parts:

\[
F^{-1}(F(E)) = E
\]

This completes the proof.

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Key Conditions and Their Significance

To understand why the theorem holds, it's essential to recognize the roles of injectivity and surjectivity:

• Injectivity ensures that different elements in \(A\) map to different elements in \(B\), which allows us to uniquely identify pre-images.


• Surjectivity of \(F|_E\) guarantees that every element in the image \(F(E)\) has a pre-image within \(E\). This is crucial because it prevents elements outside \(E\) from mapping into \(F(E)\), ensuring the inverse image doesn't include extraneous elements.

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Example to Illustrate the Theorem

Let's consider a concrete example to solidify the concepts.

Setup of the Example

• Sets:

\[ A = \{1, 2, 3, 4\} \] \[ B = \{a, b, c, d\} \]

• Function \(F : A \to B\):

\[ F(1) = a \] \[ F(2) = b \] \[ F(3) = c \] \[ F(4) = d \]

• Subset \(E \subseteq A\):

\[ E = \{2, 3\} \]

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Analysis of the Example

• Check injectivity:
• Each element of \(A\) maps to a unique element in \(B\).
• Yes, \(F\) is injective.
• Check the restriction \(F|_E : E \to F(E)\):
• \(F(2) = b\)
• \(F(3) = c\)
• \(F(E) = \{b, c\}\)
• The mapping \(F|_E\) is onto \(F(E)\) because:
• For each \(b, c \in F(E)\), there exists an element in \(E\) mapping to it.
• Yes, \(F|_E\) is onto \(F(E)\).
• Compute \(F(E)\):

\[ F(E) = \{b, c\} \]

• Find \(F^{-1}(F(E))\):

\[ F^{-1}(\{b, c\}) = \{a \in A : F(a) \in \{b, c\}\} \]

• \(F(2) = b \Rightarrow 2 \in F^{-1}(F(E))\)
• \(F(3) = c \Rightarrow 3 \in F^{-1}(F(E))\)
• \(F(1) = a \notin \{b, c\}\)
• \(F(4) = d \notin \{b, c\}\)

\[ \Rightarrow F^{-1}(F(E)) = \{2, 3\} = E \]

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Conclusion from the Example

This example perfectly illustrates the theorem:

• The function \(F\) is injective.


• Its restriction to \(E\) is onto \(F(E)\).


• The inverse image of \(F(E)\) under \(F\) equals \(E\).

This confirms the theoretical result with a tangible case.

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Implications and Applications of the Result

Understanding this property has several notable implications:

• In Set Theory and Functions:


• It confirms that for injective functions, the inverse image of the