Inverse Element of Bijection

Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ be a bijection.


Then:

$\map {f^{-1} } y = x \iff \map f x = y$

where $f^{-1}$ is the inverse mapping of $f$.


Proof

Suppose $f$ is a bijection.

Because $f^{-1}$ is a bijection from Bijection iff Inverse is Bijection, it is by definition a mapping.

The result follows directly from Image of Element under Inverse Mapping.

$\blacksquare$


Sources

  • 1965: Claude Berge and A. Ghouila-Houri: Programming, Games and Transportation Networks ... (previous) ... (next): $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 13$
  • 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.4$
  • 2008: David Joyner: Adventures in Group Theory (2nd ed.) ... (previous) ... (next): Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Definition $2.1.8$
  • 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
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