Prove the inverse relation $f^{-1}$ of a function $f : A\rightarrow B$ is a function from B to A if and only if $f$ is bijective.

Question

This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : B\to A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?

Answer 1

It's actually trivial.

The definition of $f^{-1}:B \to A$ being surjective is:

$f^{-1}:B \to A$ is surjective if for every $a \in A$ there exists a $b\in B$ so that $f^{-1}(b) = a$.

And as $f:A \to B$ is a function for every $a \in A$ there is an $f(a) \in B$.

So $f^{-1}(f(a)) = a$.

Hence $f^{-1}$ is surjective.

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how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?

$f:A \to B$ is a function so for every $a \in A$ there is exactly one distinct $(a,b) \in f$ for precisely one distinct $b\in B$. That's the definition of function.

So by fact 1) $f(a) = b$.

And by fact 2) we have $(b,a) \in f^{-1}$.

ANd by fact 1) again, that means $f^{-1}(b) =a$.

So for every $a \in A$ we have a $b \in B$ so that $f^{-1}(b) = a$.

And that is exactly the definition of surjective.