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[parent] mapping of period $n$ is a bijection (Proof)

Theorem Suppose $X$ is a set. Then a mapping $f:X\to X$ of period $n$ is a bijection.

Proof. If $n=1$ , the claim is trivial; $f$ is the identity mapping. Suppose $n=2,3,\ldots$ . Then for any $x\in X$ , we have $x=f\big(f^{n-1}(x)\big)$ , so $f$ is an surjection. To see that $f$ is a injection, suppose $f(x)=f(y)$ for some $x,y$ in $X$ . Since $f^n$ is the identity, it follows that $x=y$ . $ \Box$




"mapping of period $n$ is a bijection" is owned by Koro. [ owner history (1) ]
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Cross-references: identity, injection, surjection, identity mapping, proof, bijection, mapping, theorem
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This is version 4 of mapping of period $n$ is a bijection, born on 2003-08-01, modified 2004-03-12.
Object id is 4540, canonical name is MappingOfDegreeNIsASurjection.
Accessed 2235 times total.

Classification:
AMS MSC03E20 (Mathematical logic and foundations :: Set theory :: Other classical set theory )

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