Can we sheaf-theoretically force a violation of the continuum hypothesis in a (nice) topos which is *bicomplete*

Question

$\newcommand{\p}{\mathcal{P}}$I recently dug through the exercises and details in Mac Lane and Moerdijk's book "Sheaves in Geometry and Logic" which concern themselves with (a baby version of) Cohen forcing. Modulo a mistaken proof of the Souslin property, this was a really enjoyable and interesting read; I'm now wondering if we can even further improve this construction. The question is in the title but I'm going to give the context now:

So, I guess in our ambient copy of ZFC where we prove everything, we fix a set $B$ with $\aleph_2\preceq|B|$, define the poset $\p$ of partial functions $B\times\Bbb N\to\{0,1\}$ defined on a finite domain, ordered by $p\le q$ iff. $p$ extends $q$, and consider $\mathcal{E}:=\mathrm{Sh}_{\neg\neg}(\mathrm{Psh}(\p))=\mathrm{Sh}(\p;\text{dense topology})$. We then consider $\mathcal{F}:=\mathcal{E}/\mathfrak{U}$ where $\mathfrak{U}$ is any ultrafilter in (the complete Boolean algebra, so we can use Zorning) $\mathrm{Sub}_{\mathcal{E}}(1)$.

It can be shown that:

$(\ast)$: $\mathcal{F}$ is a well-pointed topos satisfying the (strong) axiom of choice, which has a natural numbers object $\Bbb N$ and an object $\mathscr{A}$ with strict cardinal inequalities $\Bbb N\prec\mathscr{A}\prec\mathscr{P}(\Bbb N)$.

For well-pointed topoi, strict cardinal inequality means what you would expect; there is a monomorphism but no epimorphism from the smaller thing to the greater thing.

However, I made the observation in my notes that, very plausibly (I did not bother to make a counterexample) it is the case that $\mathcal{F}$ is not complete (or cocomplete). The essential issue is that products of cardinality $\kappa$ need not exist unless $\mathfrak{U}$ is stable under intersections of $\kappa$-many objects; we don't have reason to believe, I don't think, this could be arranged for arbitrary $\kappa$ – just finite ones. Were it not for this wrinkle, I would be satisfied that $\mathcal{F}$ is a very good model of [something], for my own personal intuition. Because of course, I want to make all the constructions familiar to me and I should have arbitrary limits and colimits; I definitely don't mean anything formal when I say "very good model of something".

Is it possible to find a topos satisfying the properties listed in $(\ast)$ which is genuinely complete (and thus bicomplete)?

There's a chance this is quite subtle or maybe even impossible because somehow "complete" is with reference to small limits, and we are doing something like creating a "new" "model" of set theory (except not really, not without a lot of extra work – the exact status is unclear to me) so we should not expect completeness with respect to our starting model of sets, and maybe we should instead demand completeness with respect to limits which have a "small" shape, in the eyes of $\mathcal{F}$?

Answer 1

$\require{AMScd}$Well, I don't have a full answer since maybe there is something about "internal" completeness still to be said, but someone hinted the following to me - which more or less says the answer to my question is independent of ZFC (I think??) which is freaky but maybe unsurprising.

Theorem: a topos $\mathcal{E}$ is well-pointed and cocomplete if and only if it is equivalent to $\mathsf{Set}$, in which case the global sections functor $\Gamma=\mathcal{E}(1,-):\mathcal{E}\to\mathsf{Set}$ is a canonical choice of equivalence.

We can replace "cocomplete" with complete or bicomplete here. For complete topoi are necessarily bicomplete, and conversely if we are equivalent to $\mathsf{Set}$ we are necessarily complete as well.

Here, I include "locally small" as part of my axioms for a topos, which I think we all should do. We want subobject lattices to be... actual lattices, for instance.

Given this theorem - which I prove below - then a topos $\mathcal{F}$ satisfying the properties I desire exists if and only if $\mathsf{Set}$ itself violates the continuum hypothesis, which is independent of ZFC. Maybe there's a curious philosophical lesson here, about "completeness" being a little unnatural in that it enforces an external notion of how the universe of sets behaves onto what you'd want to be an internal property; also I suppose today I learnt that making topoi too set-like (which is what I wanted in my question!) actually forces them to be "sets".

Anyway, onto my proof. Theorem references are to "Sheaves in geometry and logic".

One direction is trivial. In the other, assume $\mathcal{E}$ to be well-pointed and cocomplete. Modulo global choice issues, we can define a coproduct functor $\mathcal{S}:\mathsf{Set}\to\mathcal{E}$ which assigns $T\mapsto\bigsqcup_{t\in T}1$ and for a function $f:T\to T'$ acts, using the universal property, via $\mathcal{S}(f)\circ i_t=i_{f(t)}:1\hookrightarrow\bigsqcup_{t'\in T'}1$. We want to show $\Gamma\circ\mathcal{S}$ and $\mathcal{S}\circ\Gamma$ are naturally isomorphic with the identity. Neither is completely trivial. We quote $\rm{VI}.2.7$, and note $\mathcal{E}$ is Boolean and two-valued. I use $i$ to denote inclusions, in particular labelled coproduct inclusions.

To see $\mathcal{S}\circ\Gamma$ is isomorphic to the identity on $\mathcal{E}$, for each $X\in\mathcal{E}$ there is an obvious map $\bigsqcup_{x\in\mathcal{E}(1,X)}1\to X$ given on each $x$th component simply by the map $x$ itself. This is clearly a natural transformation $\mathcal{S}\circ\Gamma\implies1_{\mathcal{E}}$. Now, suppose tthe global sections $x\neq x'$ of $X$. Consider the intersection of these subobjects of $X$; by two-valuedness and the fact the intersection must be a subobject of the domains, $1$, it is either $1$ or $0$. Were it $1$, that'd say (by terminality, $1\to1$ from the pullback must be the identity) $x=x'$ by commutativity of the pullback square - but this is not so. So the intersection is necessarily zero i.e. different global sections are disjoint. Now by theorem $\rm{IV}.7.6$ the arbitrary coproduct of pairwise disjoint subobjects is the supremum of those subobjects, in particular the resultant map from the coproduct is monic. That is to say, $\mathcal{S}\Gamma(X)\hookrightarrow X$ is a monomorphism.

We know $(\rm{IV}.2.2)$ that to see it is an isomorphism, it remains to demonstrate it is an epimorphism. Say $f,g:X\to Y$ are distinct arrows. Their equaliser $E\hookrightarrow X$ is then a proper subobject of $X$, so it has a nonzero complement (here we use Booleanism) $K=\neg E\subseteq X$. Since $K$ is nonzero and $\mathcal{E}$ is well-pointed, by considering the necessarily distinct (exercise) arrows $\mathrm{true}_K,\mathrm{false}_K:K\to 1\to\Omega$ it follows $K$ has at least one global section $\kappa:1\hookrightarrow K$. Composing with $i_K:K\hookrightarrow X$, we see there is a global section $x:=i_K\circ\kappa$ of $X$ which distinguishes $f$ from $g$; indeed, if $f\circ x=g\circ x$ then $x$ factors through $E\hookrightarrow X$ so that it factors through the intersection $E\cap K=E\cap\neg E=0$; but there is no arrow $1\to0$ since $\mathcal{E}$ is nondegenerate (part of the well-pointed definition), so $f\circ x\neq g\circ x$ follows. Then by the universal coproduct property $f,g$ are distinguished by $\mathcal{S}\Gamma(X)\hookrightarrow X$, as they differ on the $x$th component. This implies this arrow is an epimorphism as well as a monomorphism, hence $\mathcal{S}\Gamma(X)\cong X$ naturally in $X$.

To see $\Gamma\circ\mathcal{S}\cong\mathrm{1}_{\mathsf{Set}}$, there is first a fairly obvious and natural injection $T\hookrightarrow\mathcal{E}(1,\bigsqcup_{t\in T}1)$. To argue injectivity we have to argue that if $t\neq t'$, $i_t\neq i_{t'}$ in the coproduct. First recall from $\rm{IV}.10.4$ that the following diagrams (evident pushouts) are also pullbacks and consist entirely of monomorphisms: $$\begin{CD}0@>>>1\\@VVV@VVV\\1@>>>1\sqcup1\end{CD}\quad\quad\begin{CD}0@>>>1\sqcup1\\@VVV@VV\langle i_t,i_{t'}\rangle V\\\bigsqcup_{t''\in T\setminus\{t,t'\}}1@>>>\bigsqcup_{t''\in T}1\end{CD}$$In particular the rightmost vertical is a monomorphism and it follows, by composing this monomorphism with the lefthand pullback, that: $$\begin{CD}0@>>>1\\@VVV@VVi_tV\\1@>>i_{t'}>\bigsqcup_{t'''\in T}1\end{CD}$$So these arrows could not be equal unless $0\cong1$, which is false by assumption. So yes, it injects (and I apologise for the verbosity but it pays to be careful! I don't think this assertion needs to be true in general categories!).

It remains to check it surjects i.e. that every global section of $\mathcal{S}(T)$ takes the form $i_t$ for some $t$. Infer as a corollary of $\rm{IV}.7.2$ that pullbacks preserve coproducts. If $x:1\hookrightarrow\mathcal{S}(T)$ is given, then as of course (in a slice category) $\mathrm{id}:\mathcal{S}(T)=\mathcal{S}(T)$ can be realised as the coproduct of all the $i_t:1\hookrightarrow\mathcal{S}(T),t\in T$ so we get the following pullback diagrams: $$\begin{CD}F_t&@>>>1\\@VVV@VVi_tV\\1@>>x>\mathcal{S}(T)\end{CD}\quad\quad\begin{CD}\bigsqcup_{t\in T}F_t@>>>\mathcal{S}(T)\\@V\cong VV@VV=V\\1@>>x>\mathcal{S}(T)\end{CD}$$In any category with initial object, the arbitrary coproduct of zero is again zero, so once again we use the fact $0$ is not isomorphic with $1$ to infer not all $F_t$ can be isomorphic with zero. As monomorphisms $(x)$ pullback to monomorphisms and $\mathcal{E}$ is two-valued, each $F_t\hookrightarrow1$ is either the zero subobject or the full subobject i.e. an isomorphism. Hence at least one $F_t\to1$ is an isomorphism. But then, for this $t$, we can see $x=i_t$; $1$ is terminal, so we don't get a mysterious isomorphism $1\cong1$ which exchanges $x$ with $i_t$ - they must literally be equal, this isomorphism literally identity. So (very nonconstructively) there exists a $t\in T$ such that $x=i_t$, and this $t$ must be unique by the previous observation.

All done! With such $\mathcal{S}$ as candidate pseudoinverses, $\Gamma:\mathcal{E}\simeq\mathsf{Set}$ is an equivalence.