ii. Let f: A → B and g: B → C be invertible mappings; that is, mappings such that f-¹ and g-¹ exist. Show that (gof)-¹ = f-¹g-¹.
ii. Let f: A → B and g: B → C be invertible mappings; that is, mappings such that f-¹ and g-¹ exist. Show that (gof)-¹ = f-¹g-¹.
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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![### Invertible Mappings and Their Compositions
**Problem Statement:**
Let \( f: A \rightarrow B \) and \( g: B \rightarrow C \) be invertible mappings; that is, mappings such that \( f^{-1} \) and \( g^{-1} \) exist. Show that:
\[ (g \circ f)^{-1} = f^{-1} \circ g^{-1} \]
### Explanation:
1. **Definitions:**
- \( f \) is a function that maps elements from set \( A \) to set \( B \).
- \( g \) is a function that maps elements from set \( B \) to set \( C \).
- Both \( f \) and \( g \) are invertible, meaning they have corresponding inverse functions \( f^{-1} \) and \( g^{-1} \), respectively.
2. **Goal:**
- We need to show that the inverse of the composition \( g \circ f \) is equal to the composition of the inverses in the reverse order.
### Proof Outline:
1. Consider any element \( c \) in set \( C \). First, we apply \( g^{-1} \) to map \( c \) back to an element \( b \) in set \( B \):
\[ b = g^{-1}(c) \]
2. Next, apply \( f^{-1} \) to map \( b \) back to an element \( a \) in set \( A \):
\[ a = f^{-1}(b) = f^{-1}(g^{-1}(c)) \]
3. Therefore, for an element \( c \) in \( C \):
\[ a = (f^{-1} \circ g^{-1})(c) \]
4. By definition of inverses and composition of functions, \( (g \circ f)(a) \) should map \( a \) back to \( c \):
\[ (g \circ f)(a) = c \]
5. As a result:
\[ (g \circ f)^{-1}(c) = (f^{-1} \circ g^{-1})(c) \]
Thus, we have shown that \( (g \circ f)^{-1} = f^{-1} \circ g^{-1} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbb66fa7-7c22-4982-a22f-aaed542f65b3%2Fa2948a9c-646e-49b1-babd-93c4a89e3e9c%2Fdcwqppp_processed.png&w=3840&q=75)
Transcribed Image Text:### Invertible Mappings and Their Compositions
**Problem Statement:**
Let \( f: A \rightarrow B \) and \( g: B \rightarrow C \) be invertible mappings; that is, mappings such that \( f^{-1} \) and \( g^{-1} \) exist. Show that:
\[ (g \circ f)^{-1} = f^{-1} \circ g^{-1} \]
### Explanation:
1. **Definitions:**
- \( f \) is a function that maps elements from set \( A \) to set \( B \).
- \( g \) is a function that maps elements from set \( B \) to set \( C \).
- Both \( f \) and \( g \) are invertible, meaning they have corresponding inverse functions \( f^{-1} \) and \( g^{-1} \), respectively.
2. **Goal:**
- We need to show that the inverse of the composition \( g \circ f \) is equal to the composition of the inverses in the reverse order.
### Proof Outline:
1. Consider any element \( c \) in set \( C \). First, we apply \( g^{-1} \) to map \( c \) back to an element \( b \) in set \( B \):
\[ b = g^{-1}(c) \]
2. Next, apply \( f^{-1} \) to map \( b \) back to an element \( a \) in set \( A \):
\[ a = f^{-1}(b) = f^{-1}(g^{-1}(c)) \]
3. Therefore, for an element \( c \) in \( C \):
\[ a = (f^{-1} \circ g^{-1})(c) \]
4. By definition of inverses and composition of functions, \( (g \circ f)(a) \) should map \( a \) back to \( c \):
\[ (g \circ f)(a) = c \]
5. As a result:
\[ (g \circ f)^{-1}(c) = (f^{-1} \circ g^{-1})(c) \]
Thus, we have shown that \( (g \circ f)^{-1} = f^{-1} \circ g^{-1} \).
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