Composite of Bijections is Bijection/Proof 1
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Theorem
Let $f$ and $g$ be mappings such that $\Dom f = \Cdm g$.
Then:
- If $f$ and $g$ are both bijections, then so is $f \circ g$
where $f \circ g$ is the composite mapping of $f$ with $g$.
Proof
As every bijection is also by definition an injection, a composite of bijections is also a composite of injections.
Every composite of injections is also an injection by Composite of Injections is Injection.
As every bijection is also by definition a surjection, a composite of bijections is also a composite of surjections.
Every composite of surjections is also a surjection by Composite of Surjections is Surjection.
As a composite of bijections is therefore both an injection and a surjection, it is also a bijection.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations: $\text{(ii)}$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.10 \ (3)$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 25.2$: Some further results and examples on mappings
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions