My question is whether the following has been considered as an axiom, and if so, where I may find a discussion of it:



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> $\Xi : \prod\_{A : \mathcal{U}} \|A\| \to A$.
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For example, using this axiom, we can define the $n$-truncation of a type $A$ in the following way:


Let 



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> $P(A) :\equiv \sum\_{X : \mathcal{U}} \sum\_{\alpha : A \to X}\sum\_{\beta : X \to A} (\alpha \circ \beta = \text{id}\_B) \times (\text{is-}n\text{-type }X)$.
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We easily find that $P(A)$ is inhabited ($X :\equiv \|A\|$, $\alpha :\equiv |\cdot|\_A$, $\beta :\equiv \Xi\_A$, $\dots$):



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> p : P(A).
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Using $\Xi\_{P(A)} : \|P(A)\| \to P(A)$, we obtain a term



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> $\Xi\_{P(A)}(|p|\_{P(A)}) : P(A)$.
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My impression is that the $X$ that may be extracted from this last term behaves like the freest possible $n$-type with retraction from A on it, which in turn should behave just like $\|A\|\_n$. Is that true?



The axiom $\Xi : \prod_{A:\mathcal{U}} \|A\| \to A$ and the $n$-truncation of a type
This rule contradicts the univalence axiom.
Indeed, first note that it implies the axiom of choice:
$$ \prod\_{x : A} \| B x \| \to \| \prod\_{x : A} B x \|.$$
$\Xi$ implies that $\prod\_{x : A} \| B x \| \to \prod\_{x : A} B x$. Hence, the axiom of choice holds.


The axiom of choice implies the law of excluded middle:
$$ \prod\_{P : Prop} P \lor \neg P.$$
This implies that for every set $A$, $\| A \| \lor \neg A$. It follows that for every set $A$, $\| A + \neg A \|$, and $\Xi$ implies that
$$ \prod\_{A : Set} A + \neg A,$$
which contradicts the univalence axiom.


Also, consider the following weakening of $\Xi$:
$$\Xi' : \prod\_{A : \mathcal{U}} \| \| A \| \to A \|.$$
The law of excluded middle implies this axiom. Indeed, apply LEM to $\| A \|$. If $\| A \|$ holds, then $\| \| A \| \to A \|$ also holds. If $\|A\| \to \bot$ holds, then $\| A \| \to A$ holds.
Thus, we have the following implications:
$$ \Xi \to AC \to LEM \to \Xi'.$$


