Prove that the statements are equivalent
- For every $b \in B$, the set $f^{-1}(b)$ has exactly one element.
- $f$ is bijective.
Hint: it involves stating two facts: $f^{-1}\neq \emptyset$ and $f^{-1}(b)$ has only one element.
Thankyou in advanced for your help :)
$f^{-1}(b)\neq \emptyset$ for every $b\in B$ means $f:A\rightarrow B$ is surjective and $|f^{-1}(b)|=1$ for every $b\in B$ (i.e. every fibre contains exactly one element) even implies it is surjective and injective. Thus there exists a map $f^{-1}:B\rightarrow A$ that sends each $b\in B$ to the element in its fibre $f^{-1}(b)$ and that map obeys the equalities $$f^{-1}\circ f=id_A,f\circ f^{-1}=id_B$$ where $id_A$ and $id_B$ are the identity maps on $A$ and $B$ respectively. Thus $f$ is an isomorphism in the category of sets, i.e. a bijection.