# Composite of Bijections is Bijection/Proof 2

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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

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

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $2$: Maps and relations on sets: Corollary $2.16$