# 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