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As I understand from this question, without the axiom of choice, we can have surjective functions without right-inverse. Is that correct? Is there any such example of a surjective function where we need to use the AoC to prove the existence of the right-inverse?

Robin to Roxel
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    Could you describe your well-ordering on $\mathbb{R}$ in a bit more detail? (Hint: you can't do it in plain old ZF.) – Patrick Stevens Apr 01 '22 at 08:18
  • From your hint, of course the well-ordering is equivalent to AoC, so I removed that part from my question. – Robin to Roxel Apr 01 '22 at 08:24
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    https://math.stackexchange.com/questions/2485267/ https://math.stackexchange.com/questions/1383312/ https://math.stackexchange.com/questions/726611/ probably a few more. There's a lot more about well orders of the reals, though. – Asaf Karagila Apr 01 '22 at 08:29
  • What function exactly would be such an example? – Robin to Roxel Apr 01 '22 at 08:51
  • Mapping each sequence of reals to its range; mapping each subset of $\Bbb R$ which is well-ordered (in the natural order) to its order type. – Asaf Karagila Apr 01 '22 at 09:02
  • Is difference between highest and lowest element of the sequence meant with range or image of the sequence? – Robin to Roxel Apr 01 '22 at 09:30
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    Range as in the set of reals that appear in that sequence. A sequence, after all, is nothing but a function. – Asaf Karagila Apr 01 '22 at 09:47
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    It is not quite true that without AoC you can have surjective functions without right inverses. What is true is that you cannot always prove that a right inverse must exist. But to prove the existence of a surjective function without a right inverse, you have to go farther: you have to add an axiom that contradicts the AoC. If the AoC is known to be false, then you can use any counter-example to the AoC to construct a surjective map without right inverse. – Paul Sinclair Apr 02 '22 at 04:07

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From the comments: Let $S$ be the set of all sequences $f: \mathbb{N} \to \mathbb{R}$ and $T = \{Im(f)| f\in S\}$. Then the function $\phi: S \to T$ with $\phi(f)=Im(f)$ is surjective by definition. In order to define the right-inverse of $\phi$ for a given $t \in T$, you need to assign an $n \in \mathbb{N}$ to every real number in $t$. For this, you need the axiom of choice.

Robin to Roxel
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