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Say a function is defined and we find that it is indeed onto (surjective). Does this mean the function also has an inverse function?

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  • $\begingroup$ To have an inverse, it must be a bijection (both injective and surjective) since if it fails to be injective then it “loses” information by mapping two points into one, and there is no way of inverting this. $\endgroup$ – Jack Crawford Jul 3 at 5:31
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If $f\colon A\to B$ is onto, then there exists a function $g\colon B\to A$ such that $g\circ f=\operatorname{id}_A$ (given $b\in B$, let $g(b)$ be any $a\in A$ with $f(a)=b$, which exists per surjectivity). But this function need not be unique, nor do we have $f\circ g=\operatorname{id}_B$ (i.e., $g$ is only a left inverse), and in fact the existence of $g$ is not "constructive" in that in general it requires the Axiom of Choice.

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