# Proof

### "$\Rightarrow$"

• 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 $f(g(b))=f(a)=b.$

### "$\Leftarrow$"

• 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.

### Uniqueness of $g$

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### References

#### Bibliography

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