# Composite of Bijections is Bijection

## Theorem

Every composite of bijections is also a bijection.

That is:

- If $f$ and $g$ are both bijections, then so is $f \circ g$.

## Proof 1

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.

$\blacksquare$

## Proof 2

Let $g: X \to Y$ and $f: Y \to Z$ be bijections.

Then from Bijection iff Inverse is Bijection, both $f^{-1}$ and $g^{-1}$ are bijections.

From Inverse of Composite Relation we have that $g^{-1} \circ f^{-1}$ is the inverse of $f \circ g$.

Then:

- $\paren {f \circ g} \circ \paren {g^{-1} \circ f^{-1} } = I_Z$
- $\paren {g^{-1} \circ f^{-1} } \circ \paren {f \circ g} = I_X$

Hence the result.

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{I}$: Exercise $\text{J}$ - 1968: Ian D. Macdonald:
*The Theory of Groups*... (previous) ... (next): Appendix: Elementary set and number theory