Exercise 2: Let f: X → Y and g: Y → Z be two bijective functions. Show that (gof)-1 = ƒ-1 o g¯1. Remark: This is frequently referred to as “shoes and socks" or something similar. Let f be the action of putting on socks, and g the action of putting on shoes. Then in order to get properly dressed, one ususally does go f: you put your socks on first, and then put your shoes on. However, at the end of the day, one does the opposite to undo this: one takes off their shoes first, and then their socks. Thus, (go f)-1 = f-1og¬1. Thus, the result makes sense. Note this is false for injective functions for a trivial reason, that the functions may not be possible to compose them.

Exercise 2: Let f: X → Y and g: Y → Z be two bijective functions. Show that (gof)-1 = ƒ-1 o g¯1. Remark: This is frequently referred to as “shoes and socks" or something similar. Let f be the action of putting on socks, and g the action of putting on shoes. Then in order to get properly dressed, one ususally does go f: you put your socks on first, and then put your shoes on. However, at the end of the day, one does the opposite to undo this: one takes off their shoes first, and then their socks. Thus, (go f)-1 = f-1og¬1. Thus, the result makes sense. Note this is false for injective functions for a trivial reason, that the functions may not be possible to compose them.

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter6: Vector Spaces

Section6.2: Linear Independence, Basis, And Dimension

Problem 15EQ

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