Composite of Bijections is Bijection

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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