Yes, it is possible.

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- March 29th 2009, 06:23 PM #1

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## 1:1 and onto

Is it possible for a function to be 1:1 and not onto and is it possible for a function to be onto but not 1:1? The catch is the function is from a set A->A. I can't find a set A and a function f so f:A->A that fits these (two separate functions and sets to meet each condition). Any hints on this are greatly appreciated.

- March 29th 2009, 07:03 PM #2

- March 29th 2009, 07:11 PM #3

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- March 29th 2009, 07:37 PM #4

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