(related to Proposition: Characterization of Bijective Functions)

- Assume, the function $f:A\to B$ be bijective.
- By definition, $f$ is surjective and injective.
- Since $f$ is surjective, for every $b\in B$ there is exactly one $a\in A$ with $b=f(a).$
- Set $g:B\to A$ be the function defined by $g(b)=a$
- Obviously, the required compositions are $g(f(a))=g(b)=a$ and $f(g(b))=f(a)=b.$

- Let $g:B\to A$ be a function with the required compositions $g(f(a))=g(b)=a$ and $f(g(b))=f(a)=b$ for all $a\in A$ and $b\in B.$
- By definition, $f\circ g=id_A$ is the identity function on $A.$ Thus, $f$ is surjective.
- For any $a_1,a_2\in A$ with $f(a_1)=f(a_2)$ we have $a_1=g(f(a_1))=g(f(a_2))=a_2$. Thus, $f$ is injective.
- Since $f$ is injective and surjective, it follows that $f$ is objective.

- If $g$ and $h$ are two functions with the required properties, then because the composition of functions is associative, $$g=g\circ id_B=g\circ (f\circ h)=(g\circ f)\circ h=id_A\circ h=h.$$∎

**Modler, F.; Kreh, M.**: "Tutorium Analysis 1 und Lineare Algebra 1", Springer Spektrum, 2018, 4th Edition