sols14

# sols14 - 32 14 October 18th Isomorphisms 14.1 Let F be a...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 32 14. October 18th: Isomorphisms 14.1 . Let F be a family of sets. Prove that the isomorphism relation ∼ = defines an equivalence relation on F . That is, show that (for all sets A , B , C in F ) • A ∼ = A ; • If A ∼ = B , then B ∼ = A ; • If A ∼ = B and B ∼ = C , then A ∼ = C . Answer: • A ∼ = A because 1 A : A → A is an isomorphism. • A ∼ = B = ⇒ B ∼ = A : if f : A → B is an isomorphism, then it has an inverse function f- 1 : B → A , which is also an isomorphism. • If A ∼ = B and B ∼ = C , then there are isomorphisms f : A → B and g : B → C . The composition g ◦ f : A → C is then an isomorphism (Exercise 12.3), proving that A ∼ = C . 14.2 . A certain function f : Z → Z is defined by f ( x ) = ax + b , where a and b are integers. Assume that f is an isomorphism. What can you say about a and b ? What is the inverse of f ? Answer: If f is an isomorphism, it has an inverse g : Z → Z . Since f ◦ g = 1 Z , we must have f ◦ g ( n ) = n for all n ∈ Z , that is: n = f ◦ g ( n ) = f ( g ( n )) = ag ( n ) + b . This says that g ( n ) = n- b a . Thus a must be an integer such that ( n- b ) /a is an integer for all n . It follows that a = ± 1. Thus, either f ( n ) = n + b and f- 1 ( n ) = g ( n ) = n- b or f ( n ) =- n + b and f- 1 ( n ) = g ( n ) =- n + b . There is no restriction of what b may be. 14.3 . List all the elements of the set of bijections of A = { a, b,c, d } to itself, by indicating the image of each element. For example, a → a b → c c → b d → d denotes the bijection which swaps b and c . Next to each element, write its inverse. • Find one element s that is not the identity and such that s ◦ s is the identity. • Find one element t that is not the identity and such that t ◦ t ◦ t is the identity. • Find one element u that is not the identity, such that u ◦ u is not the identity, and such that u ◦ u ◦ u ◦ u is the identity. 14. OCTOBER 18TH: ISOMORPHISMS 33 Answer: There are 24 distinct bijections: a → a b → b c → c d → d , a → a b → b c → d d → c , a → a b → c c → b d → d , a → a b → c c → d d → b , a → a b → d c → b d → c , a → a b → d c → c d → b a → b b → a c → c d → d , a → b b → a c → d d → c , a → b b → c c → a d → d , a → b b → c c → d d → a , a → b b → d c → a d → c , a → b b → d c → c d → a a → c b → a c → b d → d ,...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

sols14 - 32 14 October 18th Isomorphisms 14.1 Let F be a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online