Background: In functorial algebraic geometry one would like to consider the category of all functors $\mathsf{CRing} \to \mathsf{Set}$ and define/characterize the category of schemes as a full subcategory. However, there are serious set-theoretic difficulties: this is not a category, it is too large. One possible solution is suggested in Demazure-Gabriel's book. They use three nested universes. I wonder if we may stay in one universe. In general, one constructs the cocompletion $\widehat{\mathcal{C}}$ of a category $\mathcal{C}$ (not assumed to be small) as the category of functors $F : \mathcal{C}^{\mathrm{op}} \to \mathsf{Set}$ which are cofinally small, i.e. the [category of elements](http://ncatlab.org/nlab/show/category+of+elements) $\int F$ is [cofinally small](http://ncatlab.org/nlab/show/cofinally+small+category). Then $\widehat{\mathcal{C}}$ is a category. We may apply this to $\mathcal{C}^{\mathrm{op}}=\mathsf{CRing}$ and develop functorial algebraic geometry inside $\widehat{\mathcal{C}}$. But the question is if the usual category of (geometric) schemes embeds into it. This leads to the following questions:


**Question.** Let $X$ be a scheme. Is there a set of commutative rings $\{A\_i\}$ with the property that for every commutative ring $A$, every morphism $\mathrm{Spec}(A) \to X$ factors through some $\mathrm{Spec}(A\_i)$?


The answer is yes if $X$ is quasi-compact quasi-separated. In fact, in that case $X(-) : \mathsf{CRing} \to \mathsf{Set}$ commutes with filtered colimits so that we may take a skeleton of all finitely generated commutative rings (which is a set). If $X$ is arbitrary, I guess that $X(-)$ commutes with $\lambda$-directed colimits for some cardinal number $\lambda$, which would answer the question. But I'm not sure about that.


**Question.** In case the answer to the first question is "Yes": Can we choose that set in such a way that for every $\mathrm{Spec}(A) \to X$ the category of morphisms $\mathrm{Spec}(A) \to \mathrm{Spec}(A\_i)$ over $X$ is connected?



Is the functor of points of a scheme cofinally small?
Maybe I'm missing something, but it seems to me like the first question is just a simple definition chase. A morphism $\mathrm{Spec}(A)\to X$ is determined by finitely many affine open subsets $\mathrm{Spec}(B\_n) \subseteq X$ and some ring maps $B\_n\to A\_{f\_n}$ to localizations of $A$ satisfying certain compatibility properties. There are not so many elements of $A$ that are needed to witness all of this. Specifically, we need to witness the $f\_n$, numerators of everything in the images of the maps $B\_n\to A\_{f\_n}$, some elements annihilated by powers of the $f\_n$ that witness that the maps $B\_n\to A\_{f\_n}$ are actually homomorphisms and are appropriately compatible, and elements of $A$ that witness that some $f\_n$ are in the ideal generated by others (and that $1$ is in the ideal generated by all of them). In total, it is clear that some subring $A\_0\subseteq A$ of cardinality at most $\aleph\_0+\sum |B\_n|$ contains all the needed witnesses, and our morphism will factor through $\mathrm{Spec}(A\_0)$.


For the second question, consider two factorizations $\mathrm{Spec}(A)\to\mathrm{Spec}(A\_i)\to X$ ($i=0,1$) where $|A\_i|<\kappa\_0$ for some cardinal $\kappa\_0$ depending only on $X$. Let $Y=\mathrm{Spec}(A\_0)\times\_X \mathrm{Spec}(A\_1)$; this may not be affine, but it will still satisfy some cardinality bound depending on $X$. Applying the previous paragraph with $Y$ in place of $X$, we can factor the map $\mathrm{Spec}(A)\to Y$ through $\mathrm{Spec}(A\_2)$, and we can bound $A\_2$ by some cardinal $\kappa\_1$ depending only on $X$. Now repeat this argument allowing $A\_0$ and $A\_1$ to have cardinality up to $\kappa\_1$, and get some new bound $\kappa\_2$ on the cardinality of $A\_2$. Repeating this inductively, we get a sequence of cardinals $\kappa\_n$; let $\kappa=\sup \kappa\_n$. Then whenever $|A\_0|<\kappa$ and $|A\_1|<\kappa$, we can find $A\_2$ with $|A\_2|<\kappa$ that connects them. (Actually, I'm pretty sure that if you choose $\kappa\_0$ in the obvious way, $\kappa\_1$ will just be the same as $\kappa\_0$, so none of this iteration is actually necessary.)


