Suppose that f : A → B is a function, viewed as a relation (set of ordered pairs). Let f* denote the “reverse" relation: f* {(b, а) € В х A|(а, b) € f} For example, if f : {1,2, 3} → {*, ♡, A} is the function f = {(1, *), (2, ♡), (3, *)}, then f* = {(*, 1), (♡, 2), (*, 3)}. We proved in class that if f is a bijection, then f* is also a bijection (from B back onto A). In general, f* need not even be a function, as we can see from the example above. In fact, f* is a function if and only if f is a bijection, as you will prove below. (a) Prove that if a function f : A → B is not a bijection, then f* is not a function. (To do this, prove that if f is not an injection, then f* is not a function, and also prove that if f is not a surjection, then f* is not a function.) (b) Consider the function f : {1,2,3} function is injective, but not surjective. + {1, 2, 3, 4, 5} defined the f(n) = n + 1. Observe that this Define a surjection g : {1,2, 3, 4, 5} → {1,2,3} such that g Ɔ f*. (In class we proved such a surjection always exists.) (c) Consider the function f : Z → NU{0} defined by f(n) = |n|. Observe that this function is surjective, but not injective. Define an injection g : N → Z such that g C f*. (In class we proved such an injection always exists.)
Suppose that f : A → B is a function, viewed as a relation (set of ordered pairs). Let f* denote the “reverse" relation: f* {(b, а) € В х A|(а, b) € f} For example, if f : {1,2, 3} → {*, ♡, A} is the function f = {(1, *), (2, ♡), (3, *)}, then f* = {(*, 1), (♡, 2), (*, 3)}. We proved in class that if f is a bijection, then f* is also a bijection (from B back onto A). In general, f* need not even be a function, as we can see from the example above. In fact, f* is a function if and only if f is a bijection, as you will prove below. (a) Prove that if a function f : A → B is not a bijection, then f* is not a function. (To do this, prove that if f is not an injection, then f* is not a function, and also prove that if f is not a surjection, then f* is not a function.) (b) Consider the function f : {1,2,3} function is injective, but not surjective. + {1, 2, 3, 4, 5} defined the f(n) = n + 1. Observe that this Define a surjection g : {1,2, 3, 4, 5} → {1,2,3} such that g Ɔ f*. (In class we proved such a surjection always exists.) (c) Consider the function f : Z → NU{0} defined by f(n) = |n|. Observe that this function is surjective, but not injective. Define an injection g : N → Z such that g C f*. (In class we proved such an injection always exists.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1
Transcribed Image Text:Suppose that f : A → B is a function, viewed as a relation (set of ordered pairs). Let f* denote the
“reverse" relation: f*
{(b, а) € В х A|(а, b) € f}
For example, if f : {1,2, 3} → {*, ♡, A} is the function
f = {(1, *), (2, ♡), (3, +)},
%3|
then
f* = {(*, 1), (♡, 2), (*, 3)}.
We proved in class that if f is a bijection, then f* is also a bijection (from B back onto A). In general,
f* need not even be a function, as we can see from the example above. In fact, f* is a function if and
only if f is a bijection, as you will prove below.
(a) Prove that if a function f : A → B is not a bijection, then f* is not a function. (To do this, prove
that if f is not an injection, then f* is not a function, and also prove that if f is not a surjection,
then f* is not a function.)
(b) Consider the function f : {1,2,3}
function is injective, but not surjective.
+ {1, 2, 3, 4, 5} defined the f(n)
= n + 1. Observe that this
Define a surjection g : {1,2,3, 4, 5} → {1,2,3} such that g 2 f*. (In class we proved such a
surjection always exists.)
(c) Consider the function f : Z → NU {0} defined by f(n) = |n|. Observe that this function is
surjective, but not injective.
Define an injection g : N → Z such that g C f*. (In class we proved such an injection always
exists.)
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