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2. We already saw that the composition of bijections are bijections. In this exercise we will analyze the behavior of injections and surjections with respect to composition. (a) Show that if f: A → B and g: B → C are injections, then go f: A→ C is also injective. (b) Show that if f: A → B and g: B → C are surjections, then go f: A → C is also surjective. (c) Show that if f: A → B and g: B → C are function such that go f: A → C is a bijection, then f is injective and g is surjective. (Hint: You could either use the definitions of injections, surjections, and bijections or the fact that the injections are the monomorphisms, the surjections are the epimorphisms, and the bijections are the isomorphisms.) The pigeonhole principle says that there cannot be any injective function f: A → B if A and B are finite sets with cardinalities |A| = m and |B| = n so that n < m. It receives its name from this motivating example: if a flock of pigeons goes to nest on a dovecote containing less pigeonholes than pigeons are in the flock, then we can be sure that at least one pigeonhole will shelter at least two pigeons. This is a very intuitive principle, and its proof is a consequence of Lemma 13 of Chapter 10. Indeed, remember that we proved that if n < m then there is no injective function f: m →n.

2. We already saw that the composition of bijections are bijections. In this exercise we will analyze the behavior of injections and surjections with respect to composition. (a) Show that if f: A → B and g: B → C are injections, then go f: A→ C is also injective. (b) Show that if f: A → B and g: B → C are surjections, then go f: A → C is also surjective. (c) Show that if f: A → B and g: B → C are function such that go f: A → C is a bijection, then f is injective and g is surjective. (Hint: You could either use the definitions of injections, surjections, and bijections or the fact that the injections are the monomorphisms, the surjections are the epimorphisms, and the bijections are the isomorphisms.) The pigeonhole principle says that there cannot be any injective function f: A → B if A and B are finite sets with cardinalities |A| = m and |B| = n so that n < m. It receives its name from this motivating example: if a flock of pigeons goes to nest on a dovecote containing less pigeonholes than pigeons are in the flock, then we can be sure that at least one pigeonhole will shelter at least two pigeons. This is a very intuitive principle, and its proof is a consequence of Lemma 13 of Chapter 10. Indeed, remember that we proved that if n < m then there is no injective function f: m →n.

Elementary Geometry for College Students
Elementary Geometry for College Students
6th Edition
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Daniel C. Alexander, Geralyn M. Koeberlein
Chapter6: Circles
Section6.CT: Test
Problem 11CT: aIf HP=4, PJ=5, and PM=2, find LP. _ bIf HP=x+1, PJ=x1, LP=8, and PM=3, find x. _
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2. We already saw that the composition of bijections are bijections. In this exercise we will analyze the behavior of injections and surjections with respect to composition. (a) Show that if f: A → B and g: B → C are injections, then go f: A → C is also injective. (b) Show that if ƒ: A → B and g: B → C are surjections, then go f: A → C is also surjective. (c) Show that if f: A → B and g: B → C are function such that go f: A → C is a bijection, then f is injective and g is surjective. (Hint: You could either use the definitions of injections, surjections, and bijections or the fact that the injections are the monomorphisms, the surjections are the epimorphisms, and the bijections are the isomorphisms.) The pigeonhole principle says that there cannot be any injective function ƒ: A → B if A and B are finite sets with cardinalities |A| = m and |B| = n so that n < m. It receives its name from this motivating example: if a flock of pigeons goes to nest on a dovecote containing less pigeonholes than pigeons are in the flock, then we can be sure that at least one pigeonhole will shelter at least two pigeons. This is a very intuitive principle, and its proof is a consequence of Lemma 13 of Chapter 10. Indeed, remember that we proved that if n <m then there is no injective function f: m →n.
Transcribed Image Text:2. We already saw that the composition of bijections are bijections. In this exercise we will analyze the behavior of injections and surjections with respect to composition. (a) Show that if f: A → B and g: B → C are injections, then go f: A → C is also injective. (b) Show that if ƒ: A → B and g: B → C are surjections, then go f: A → C is also surjective. (c) Show that if f: A → B and g: B → C are function such that go f: A → C is a bijection, then f is injective and g is surjective. (Hint: You could either use the definitions of injections, surjections, and bijections or the fact that the injections are the monomorphisms, the surjections are the epimorphisms, and the bijections are the isomorphisms.) The pigeonhole principle says that there cannot be any injective function ƒ: A → B if A and B are finite sets with cardinalities |A| = m and |B| = n so that n < m. It receives its name from this motivating example: if a flock of pigeons goes to nest on a dovecote containing less pigeonholes than pigeons are in the flock, then we can be sure that at least one pigeonhole will shelter at least two pigeons. This is a very intuitive principle, and its proof is a consequence of Lemma 13 of Chapter 10. Indeed, remember that we proved that if n <m then there is no injective function f: m →n.
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