Inverse of Bijection is Bijection

Theorem

Let $f: S \to T$ be a bijection in the sense that:

$(1): \quad f$ is an injection
$(2): \quad f$ is a surjection.


Then the inverse $f^{-1}$ of $f$ is itself a bijection by the same definition.


Proof

Let $f$ be both an injection and a surjection.

From Mapping is Injection and Surjection iff Inverse is Mapping it follows that its inverse $f^{-1}$ is a mapping.


From Inverse of Inverse Relation:

$\paren {f^{-1} }^{-1} = f$

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

But then $f$, being a bijection, is by definition a mapping.

So from Mapping is Injection and Surjection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection.

$\blacksquare$


Also see


Sources

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  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.3$. Injective, surjective, bijective; inverse mappings
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.5$
  • 1968: Ian D. Macdonald: The Theory of Groups ... (previous) ... (next): Appendix: Elementary set and number theory
  • 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.11$: Relations: Theorem $11.9$
  • 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.4$: Functions: Problem Set $\text{A}.4$: $25$
  • 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $3$
  • 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.6$: Functions
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  • 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.4$
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