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Chapter 1: Problem 62

Show that if \(S\) and \(T\) are finite sets of the same size, then a function \(f\) from \(S\) to \(T\) is one-to-one if and only if it is onto. (h)

Short Answer

Expert verified
If \(S\) and \(T\) are finite sets of the same size, a function \(f: S \rightarrow T\) is one-to-one if and only if it is onto.

Step by step solution

01

- Understanding the problem

Given two finite sets, \(S\) and \(T\), of the same size, it is required to show that a function \(f: S \rightarrow T\) is one-to-one (injective) if and only if it is onto (surjective).
02

- Define injective and surjective

A function \(f\) is one-to-one (injective) if \(f(a) = f(b)\) implies \(a = b\). A function \(f\) is onto (surjective) if for every \(t \in T\), there exists an \(s \in S\) such that \(f(s) = t\).
03

- Assume \(f\) is injective

Assume \(f\) is injective. Since \(S\) and \(T\) have the same size, every element in \(S\) maps to a unique element in \(T\). Therefore, no two distinct elements in \(S\) map to the same element in \(T\).
04

- Prove \(f\) is onto

Because \(S\) and \(T\) have the same number of elements and \(f\) is injective, every element of \(T\) has to be mapped to by some element of \(S\). Therefore, \(f\) is surjective.
05

- Assume \(f\) is surjective

Now, assume \(f\) is surjective. Since \(f\) maps every element in \(S\) to an element in \(T\), and \(S\) and \(T\) are of the same size, each element of \(T\) must be hit by exactly one element of \(S\).
06

- Prove \(f\) is injective

Because every element of \(T\) is hit by exactly one element of \(S\), \(f\) must be injective because no two different elements in \(S\) can map to the same element in \(T\).
07

- Conclude the proof

Since we have shown that \(f\) being injective implies it is surjective, and \(f\) being surjective implies it is injective, the function \(f\) is one-to-one if and only if it is onto.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Function
An injective function, also known as a one-to-one function, is a type of function where each element in the domain maps to a distinct and unique element in the codomain. This means that no two different elements in the domain can map to the same element in the codomain.
For example, consider the function \( f: \{1, 2, 3\} \rightarrow \{a, b, c\} \) defined by \( f(1) = a \), \( f(2) = b \), and \( f(3) = c \). This function is injective because each element from the set \( \{1, 2, 3\} \) maps to a unique and different element from the set \( \{a, b, c\} \).

In mathematical terms, a function \( f: A \rightarrow B \) is injective if \( f(a_1) = f(a_2) \) implies \( a_1 = a_2 \) for all \( a_1, a_2 \) in the domain \( A \). Injectivity ensures that there are no duplicates in the mapping process.
Surjective Function
A surjective function, or an onto function, is a function where every element in the codomain is mapped to by at least one element in the domain. This means that for every element in the codomain, there is some element in the domain that maps to it.
For instance, if we have a function \( g: \{1, 2, 3\} \rightarrow \{a, b\} \) defined by \( g(1) = a \), \( g(2) = b \), and \( g(3) = a \), this function is surjective because every element in the codomain \( \{a, b\} \) is mapped to by at least one element in the domain \( \{1, 2, 3\} \).

In formal terms, a function \( g: A \rightarrow B \) is surjective if for every \( b \) in the codomain \( B \), there exists some \( a \) in the domain \( A \) such that \( g(a) = b \). Surjectivity guarantees that the codomain is completely covered by the function.
Bijection
A bijection, or a bijective function, is a function that is both injective and surjective. This means that every element in the domain maps to a unique element in the codomain (injective), and every element in the codomain is mapped to by some element in the domain (surjective).
For example, the function \( h: \{1, 2, 3\} \rightarrow \{a, b, c\} \) defined by \( h(1) = a \), \( h(2) = b \), and \( h(3) = c \) is bijective. Every element in the domain \( \{1, 2, 3\} \) maps to a unique element in the codomain \( \{a, b, c\} \) and every element in the codomain is covered.

In mathematical language, a function \( h: A \rightarrow B \) is bijective if it is both injective and surjective. This means that for each \( b \) in \( B \), there is exactly one \( a \) in \( A \) such that \( h(a) = b \). Bijections are important because they establish a one-to-one correspondence between the domain and codomain, indicating that the two sets have the same cardinality.

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Most popular questions from this chapter

A tennis club has \(2 n\) members. We want to pair up the members by twos for singles matches. (a) In how many ways may we pair up all the members of the club? (Hint: consider the cases of \(2,4,\) and 6 members.) (h) (b) Suppose that in addition to specifying who plays whom, for each pairing we say who serves first. Now in how many ways may we specify our pairs?

Show that if we have a function from a set of size \(n\) to a set of size less than \(n,\) then \(f\) is not one-to-one. \((\mathrm{h})\)

Problem 47 has a geometric interpretation in a coordinate plane. A lattice path in the plane is a "curve" made up of line segments that either go from a point \((i, j)\) to the point \((i+1, j)\) or from a point \((i, j)\) to the point \((i, j+1),\) where \(i\) and \(j\) are integers. (Thus lattice paths always move either up or to the right.) The length of the path is the number of such line segments. (a) What is the length of a lattice path from (0,0) to \((m, n) ?\) (b) How many such lattice paths of that length are there? (h) (c) How many lattice paths are there from \((i, j)\) to \((m, n),\) assuming \(i, j,\) \(m,\) and \(n\) are integers? (n)

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Now some number \(n\) of schools are going to send their baseball teams to a tournament, and each team must play each other team exactly once. Let us think of the teams as numbered 1 through \(n\). (a) How many games does team 1 have to play in? (b) How many games, other than the one with team \(1,\) does team two have to play in? (c) How many games, other than those with the first \(i-1\) teams, does team \(i\) have to play in? (d) In terms of your answers to the previous parts of this problem, what is the total number of games that must be played?
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