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Exercise 5.1.5
If and if , then there is a mapping of onto . We later show, with the help of the Axiom of Choice, that the converse is also true: If there is a mapping of onto , then .
Answers
Proof. Suppose that for sets and where . Then there is an . There is also an injective so that is a function from . So let be a mapping from to defined by
To show that is onto consider any and let . Thus so that
Hence is onto since was arbitrary. □