Which models of set theory are locally presentable?

For the purposes of this question, let me fix a "true" universe of sets, which I will call the "true sets". Recall that a category is locally presentable if it is cocomplete and accessible. Both these notions involve sets --- I will interpret them in terms of the true sets. Indeed, the word "category" involves sets (there should be a set of morphisms between any two objects) and again here I mean true sets.

Now choose some other model of set theory, which I will call the "fake sets". I presume that the fake sets assemble into a category, at least if the fake model is built "internally" to the true model.

Under what circumstances is the category of fake sets locally presentable?

• Is there a purely set-theoretic manner of asking the question? – Joel David Hamkins Feb 15 '16 at 20:21
• @JoelDavidHamkins No, because it's not a perfectly well-posed question, because I don't really know what a "model of sets" is. But here's a related question. If I'm not mistaken, among the universe U of sets, there some but not all of the sets are "constructible", which I think is called V. Is there a functor from U to V that is the identity on finite sets and takes colimits to colimits? It should be a sort of "rounding up" functor. – Theo Johnson-Freyd Feb 16 '16 at 2:18
• Gödel's constructible universe is usually denoted by L, and the full set-theoretic universe is usually denoted by V. If these are different, then you can't map V to L with a functor that is the identity on finite sets, since not all finite sets will be in L. If x is not in L, then {x} is a finite set that is not in L. – Joel David Hamkins Feb 16 '16 at 2:21
• If what you want is an $\in$-embedding from $V$ to $L$, this is related to my question mathoverflow.net/q/101821/1946, which is still open, but in joint work with Woodin, Magidor and others, we have a bunch of partial results. – Joel David Hamkins Feb 16 '16 at 2:25
• Ah, sorry for the notation! By "is the identity on finite sets" I mean something less strict than it sounds --- perhaps I should have said "is the identity on finite cardinalities". – Theo Johnson-Freyd Feb 16 '16 at 22:30

• I suppose it was clear what you had meant, although I find that usage to be sloppy. If $B$ is a Boolean algebra, then a $B$-valued model (in a language) is a structure with a domain of names, and an assignment of instances of those names under the fundamental relations to truth values in $B$. If $V$ is the set-theoretic universe, then $V^B$ is the class of $B$-names in the forcing sense, and this is what you meant. But there are many other $B$-valued models, for example, $W^B$ for various inner models $W$, or $V^C$ for any complete subalgebra $C$ of $B$, or inner models of $V^B$, etc. – Joel David Hamkins Feb 16 '16 at 13:45