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.
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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Transcribed Image Text: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.
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