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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, bijection, mapping
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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 1711 times total.

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AMS MSC: 03E20 (Mathematical logic and foundations :: Set theory :: Other classical set theory )
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