mapping of period is a bijection
Proof. If , the claim is trivial; is the identity mapping. Suppose . Then for any , we have , so is an surjection. To see that is a injection, suppose for some in . Since is the identity, it follows that .
|Title||mapping of period is a bijection|
|Date of creation||2013-03-22 13:48:57|
|Last modified on||2013-03-22 13:48:57|
|Last modified by||Koro (127)|