Let f: X→ Y be a map between the sets X and Y. (a) Show that f is injective if and only if there exists a map g: Y → X so that go f = idx. (b) Show that f is surjective if and only if there exists a map h: Y → X so that foh = idy.

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**Title: Functions and Their Properties**

**Introduction:**
In mathematics, understanding the properties of functions is essential. Here, we explore the conditions under which a map between two sets, \(X\) and \(Y\), is injective or surjective.

**Content:**

**Let \( f: X \to Y \) be a map between the sets \(X\) and \(Y\).**

**(a) Injectivity:**
- *Definition:* A function \(f\) is injective (or one-to-one) if distinct elements in the domain map to distinct elements in the codomain. In other words, if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\).
- *Statement:* Show that \( f \) is injective if and only if there exists a map \( g: Y \to X \) such that \( g \circ f = \text{id}_X \). 
- *Explanation:* Here, \(\text{id}_X\) denotes the identity map on \(X\), which maps every element of \(X\) to itself. The condition \( g \circ f = \text{id}_X \) means that applying \( f \) followed by \( g \) brings any element back to its original value in \(X\).

**(b) Surjectivity:**
- *Definition:* A function \(f\) is surjective (or onto) if every element in the codomain \(Y\) is mapped to by at least one element in the domain \(X\).
- *Statement:* Show that \( f \) is surjective if and only if there exists a map \( h: Y \to X \) such that \( f \circ h = \text{id}_Y \).
- *Explanation:* Here, \(\text{id}_Y\) denotes the identity map on \(Y\). The condition \( f \circ h = \text{id}_Y \) ensures that every element in \(Y\) is the image of some element from \(X\).

**Conclusion:**
Understanding these properties helps in analyzing the behavior of functions and their inverses. The existence of mappings \(g\) and \(h\) illustrating these properties is a cornerstone in mathematical analysis and set theory.
Transcribed Image Text:**Title: Functions and Their Properties** **Introduction:** In mathematics, understanding the properties of functions is essential. Here, we explore the conditions under which a map between two sets, \(X\) and \(Y\), is injective or surjective. **Content:** **Let \( f: X \to Y \) be a map between the sets \(X\) and \(Y\).** **(a) Injectivity:** - *Definition:* A function \(f\) is injective (or one-to-one) if distinct elements in the domain map to distinct elements in the codomain. In other words, if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\). - *Statement:* Show that \( f \) is injective if and only if there exists a map \( g: Y \to X \) such that \( g \circ f = \text{id}_X \). - *Explanation:* Here, \(\text{id}_X\) denotes the identity map on \(X\), which maps every element of \(X\) to itself. The condition \( g \circ f = \text{id}_X \) means that applying \( f \) followed by \( g \) brings any element back to its original value in \(X\). **(b) Surjectivity:** - *Definition:* A function \(f\) is surjective (or onto) if every element in the codomain \(Y\) is mapped to by at least one element in the domain \(X\). - *Statement:* Show that \( f \) is surjective if and only if there exists a map \( h: Y \to X \) such that \( f \circ h = \text{id}_Y \). - *Explanation:* Here, \(\text{id}_Y\) denotes the identity map on \(Y\). The condition \( f \circ h = \text{id}_Y \) ensures that every element in \(Y\) is the image of some element from \(X\). **Conclusion:** Understanding these properties helps in analyzing the behavior of functions and their inverses. The existence of mappings \(g\) and \(h\) illustrating these properties is a cornerstone in mathematical analysis and set theory.
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