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One version of the Axiom of Collection says that any surjection $A\to B$ from a class $A$ to a set $B$ is factored through by some surjection $C\to B$ where $C$ is a set.

Note that assuming $B$ is a set, the axiom of replacement ensures that $C$ is a set if and only if each fiber of $C\to B$ is a set (i.e. the map $C\to B$ is "small" in the sense of algebraic set theory). Thus it is natural to consider the following "class-collection axiom": any surjection $A\to B$ of classes is factored through by some surjection $C\to B$ whose fibers are all sets.

Has this "class-collection axiom" been studied at all? Assuming classical logic, it seems to follow from the axiom of foundation in the same way that collection follows from replacement and foundation: let the fiber of $C$ over $b\in B$ consist of all those $a\in A$ lying over $b$ and of minimal rank. Can it be proven in any intuitionistic context?

Mike Shulman
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  • I am pretty sure I do not understand the question. However, I feel that "Scott's trick" (pick a representative of minimal rank out of each equivalence class) means that the answer is no for set theories with Foundation. You might look at those texts (Aczel, maybe?) whose authors work without Foundation. Your question may be of some interest there. Gerhard "Ask Me About System Design" Paseman, 2011.10.11 – Gerhard Paseman Oct 12 '11 at 05:34
  • I haven't found any reference to such a thing in the anti-foundation literature that I've looked at. – Mike Shulman Oct 12 '11 at 16:28
  • Also, can you say what you don't understand about the question, so that I can attempt to clarify it? – Mike Shulman Oct 12 '11 at 16:28
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    Mike, I wonder if you have tried to prove that this holds in any known models of intuitionistic set theory? E.g., I would be tempted to look at these models in categories of ideals that have been considered in the algebraic set theory literature (such models certainly satisfy collection). – Michael A Warren Oct 13 '11 at 13:26
  • If you are willing to cater to my specific ignorance, sure. I essentially want undergraduate-level explanations of fiber and small, something like "fiber is like the inverse image of a point; in relational terms, fiber(b) is like the class C of sets c such that the substitution phi(c,b) holds for an appropriate formula phi which might describe a mapping between classes." Similarly "In algebraic set theory, small means not large, so think of small as meaning no bigger than the size of some unfixed big set." Sloppy, but helpful. Gerhard "Coffee First, Then Set Theory" Paseman, 2011.10.13. – Gerhard Paseman Oct 13 '11 at 17:31
  • Also, an example pitched to an undergraduate level might help. Presenting actual statements of replacement, collection, and your version in slightly more formal terms might help, e.g. "Lets pick the language of NBG so we can talk about classes; I want to know more about the schema: for any first order formula phi(x,y) that has this property , there is a formula psi(x,y) that and for which we assert is a set for all sets b." Gerhard "Ask Me About System Design" Paseman, 2011.10.13 – Gerhard Paseman Oct 13 '11 at 17:40
  • Michael Warren - no, I haven't, but that's a good suggestion. You're undoubtedly more familiar with such models than I am... (-: – Mike Shulman Oct 14 '11 at 20:51
  • @Gerhard Paseman: Given classes $\chi,\phi$ such that $\forall x (\chi(x) \to \exists y \phi(x,y))$ then there is a class $\psi \subseteq \phi$ such that $\forall x (\chi(x) \to \exists y \psi(x,y))$ and $\forall x (\chi(x) \to \exists s \forall y (y \in s \leftrightarrow \psi(x,y)))$. Scott's trick is non-constructive, but (set-)collection is constructive, so he is proposing a class-collection axiom as a constructive substitute for Scott's trick. (@Mike Shulman: is that right?) Interesting proposal, no idea if it works. – Daniel Mehkeri Oct 15 '11 at 01:35
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    Hmm, assuming my interpretation is right, then what happens for the specific case $\chi(x) :\leftrightarrow \exists y \phi(x,y)$ ? The existence hypothesis is trivial and we have no useful witnessing information on the LHS to work with, unlike in the set-collection case. – Daniel Mehkeri Oct 15 '11 at 01:55
  • I don't have an answer, just a comment about the first paragraph of the question. Unless I'm mistaken the axiom stated there follows from global choice: take $C = B$, take the map $C \to B$ to be the identity, and take the map $C \to A$ to be a right inverse of the surjection $A \to B$. So it cannot imply the axiom of replacement as it is usually formulated. Should the map $C \to A$ be required to be a set? – Trevor Wilson Oct 16 '11 at 00:23

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Maybe in

A. Joyal, I. Moerdijk: A completeness theorem for open maps, Annals of Pure and Applied Logic 70 (1994) 51-86

an axiom related to this occurs - but its study is not the main purpose of this paper. So I'm looking for a more extensive treatment of it myself.

Stephan Alexander Spahn
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