Exercise 5.1.5

If | A | | B | and if A , then there is a mapping of B onto A . We later show, with the help of the Axiom of Choice, that the converse is also true: If there is a mapping of B onto A , then | A | | B | .

Answers

Proof. Suppose that | A | | B | for sets A and B where A . Then there is an a A . There is also an injective f : A B so that f 1 is a function from ran ( f ) A . So let g be a mapping from B to A defined by

g ( b ) = { f 1 ( b ) b ran ( f ) a b ran ( f ) .

To show that g is onto consider any x A and let b = f ( a ) . Thus b ran ( f ) so that

g ( b ) = f 1 ( b ) = f 1 ( f ( a ) ) = a .

Hence g is onto since a was arbitrary. □

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2024-07-15 11:42
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