[PLEASE SEE EDITS AT BOTTOM OF QUESTION]


Consider the following set-theoretic axiom:



> 
> For each set $X$ there exists a set-indexed collection $\{C\_i \to X\}\_{i\in I\_X}$ of surjections such that for every surjection $Z\to X$ there is a map $C\_i\to Z$ for some $i$ such that the obvious triangle commutes.
> 
> 
> 


This is known as [WISC](http://ncatlab.org/nlab/show/WISC) (Weakly Initial Set of Covers), and can be interpreted as saying Choice fails to hold in at most a 'small' way. It is clearly implied by AC, and I'm willing to bet that it is independent of other usual set-theoretic axioms (ZF, say). WISC is implied by [COSHEP](http://ncatlab.org/nlab/show/COSHEP) (take $I\_X$ to be a singleton for all $X$), [SVC](http://ncatlab.org/nlab/show/small+violations+of+choice) and [AMC](http://ncatlab.org/nlab/show/axiom+of+multiple+choice).


My questions are these:


* Does anyone know of a weaker choice principle? (Edit: a global choice principle, or at least one for a sizable collection of sets, like all elements $\bigcup\_n \mathcal{P}(\mathbb{R}^n)$)


and


* In which popular/common models of set theory does WISC hold? That is, aside from the ones listed at the linked page above (which are particularly category theory-oriented).


(As a bonus question: Come up with a model of ZF that violates WISC or prove we can use forcing to construct one)




---


Edit: There is also the axiom WISC${}\_\kappa$, where we require the set $I\_X$ to be bounded by some cardinal $\kappa$ (either less than or at most). This is perhaps more interesting than the unbounded case, especially in topological applications.




---


Edit2: Benno van den Berg has now shown that [Gitik's model of ZF](https://doi.org/10.1007/BF02760939) (Israel J. Math 1980) violates WISC. This model, which relies on the consistency of a large cardinal assumption (that is, the existence of an unbounded collection of strongly compact cardinals), has the property that only $\aleph\_0$ is a regular cardinal. What [Benno showed](https://staff.fnwi.uva.nl/b.vandenberg3/papers/WISC.pdf) was that ZF+WISC implies the existence of an unbounded collection of regular cardinals. Now one can clearly ask (thanks to godelian in the comments) whether weaker large cardinal assumptions suffice. One would only need to find a model of ZF in which there is only a bounded collection of regular cardinals. This to me sounds reasonable.



On a weak choice principle
Since [Benno van den Berg's argument](https://staff.fnwi.uva.nl/b.vandenberg3/papers/WISC.pdf) is cast in category theoretic language, here is a translation (and slight simplification) for the benefit of set theorists who may be unfamiliar with the terminology. First note that we may assume that the domains $C\_i$ in the statement of WISC are all the same (see note at end). Then, since the collection of all surjections $C \to X$ forms a set, a simplified form of WISC is the following:



> 
> For every set $X$ there is a set $C$ such that for every surjection $q:Y \to X$ there is a map $s:C \to Y$ such that $q \circ s: C \to X$ is a surjection.
> 
> 
> 


Assuming WISC holds at $\omega$, we show that there is an ordinal of uncountable cofinality, the argument easily generalizes to larger cofinalities. Let $C$ be as in the simplified form of WISC for $X = \omega$.


Let $\mathcal{W}$ be the set of all wellfounded subtrees of $C^{<\omega}$. Each such tree $T$ has a rank $\mathrm{rk}(T) = \mathrm{rk}\_T(\langle\rangle)$, where $\mathrm{rk}\_T:T \to \mathrm{Ord}$ is defined by the recursive formula $$\mathrm{rk}\_T(t) = \sup\lbrace \mathrm{rk}\_T(t^\frown\langle x\rangle)+1 : x \in C \land t^\frown \langle x\rangle \in T \rbrace$$ for each $t \in T$. (Note that $\sup \varnothing = 0$ so the leaves of $T$ have rank $0$.) It is straightforward to check that $$\alpha = \lbrace\mathrm{rk}(T) : T \in \mathcal{W}\rbrace$$ is an initial ordinal. We claim that $\alpha$ is a limit ordinal of uncountable cofinality.


We cannot have $\alpha = 0$ since $\mathcal{W} \neq \varnothing$. We also cannot have $\alpha = \beta+1$ for if $\mathrm{rk}(T) = \beta$ then $$T' = \lbrace\langle\rangle\rbrace\cup\lbrace \langle x\rangle^\frown t : x \in C \land t \in T\rbrace$$ is an element of $\mathcal{W}$ with $\mathrm{rk}(T') = \beta + 1 = \alpha$. So $\alpha$ must be a limit ordinal.


To see that $\alpha$ must have uncountable cofinality, suppose instead that $\alpha = \sup\_{n \lt \omega} \alpha\_n$ where $\langle \alpha\_n \rangle\_{n \lt \omega}$ is an increasing sequence of ordinals with $\alpha\_0 = 0$. Define $q:\mathcal{W}\to\omega$ by $$q(T) = \max\lbrace n \in \omega : \mathrm{rk}(T) \geq \alpha\_n \rbrace.$$ Since this is a surjection, there are a surjection $p:C \to \omega$ and a sequence $\langle T\_x \rangle\_{x \in C}$ of elements of $\mathcal{W}$ such that $$\alpha\_{p(x)} \leq \mathrm{rk}(T\_x) \lt \alpha\_{p(x)+1}$$ for all $x \in C$. But then the tree $$T = \lbrace\langle\rangle\rbrace\cup\lbrace \langle x\rangle^\frown t : x \in C \land t \in T\_x \rbrace$$ is an element of $\mathcal{W}$ with $\mathrm{rk}(T) = \alpha$.




---


Since this is not immediately obvious, here is a detailed explanation why the domains $C\_i$ in the original statement of WISC can be assumed to be the same. The argument works in Set and any other [well-pointed](http://ncatlab.org/nlab/show/well-pointed+topos) Boolean topos.


We may assume $X \neq \varnothing$. Pick $x\_0 \in X$ once and for all. Suppose $p\_i:C\_i \to X$, $i \in I$, is a set-indexed family of surjections as in the statement of WISC. Let $C = \bigcup\_{i \in I} C\_i$ and let $\bar{p}\_i:C\to X$ be the extension of $p\_i$ with $\bar{p}\_i(a) = x\_0$ for all $a \in C - C\_i$. I claim that the surjections $\bar{p}\_i:C\to X$, $i \in I$, are also as required for WISC.


Suppose $q:Y \to X$ is a surjection. By hypothesis there are an $i \in I$ and a map $s:C\_i \to Y$ such that $q \circ s = p\_i$. Extend $s:C\_i \to Y$ to $\bar{s}:C \to Y$ by defining $\bar{s}(a) = y\_0$ for all $a \in C - C\_i$ where $y\_0$ is an element of $q^{-1}(x\_0)$. Then $q \circ \bar{s} = \bar{p}\_i$, as required.


