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The concept of a subobject classifier is of course standard and ubiquitous. But is there any nontrivial example of an unrestricted slice classifier?

Specifically, what I mean by this is, is there any non-preorder category with pullbacks with a morphism m into an object X such that ALL other morphisms can be taken as a pullback of m along some morphism into X? And, if so, is it even possible to have furthermore that parallel morphisms from any object Y into X are equal just in case the pullbacks of m along them are isomorphic as objects of the slice category over Y?

Naturally, if we demand further structure on the category (e.g., local cartesian closure), this becomes impossible by Cantor type arguments in its internal logic, but if we only demand pullbacks, can it be done?

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  • $\begingroup$ Charming question, this. You might as well consider the non-preorder case, since the preorder case degenerates into triviality: a preorder with this property is equivalent to the terminal category. Hope to get some time to think about this question more -- it feels as though the domain of $m$ ought to carry a category object structure with the conceit of internalizing the entire ambient category. Close to the razor's edge of paradox! $\endgroup$
    – Todd Trimble
    Commented Nov 11, 2010 at 22:00
  • $\begingroup$ "You might as well consider the non-preorder case, since the preorder case degenerates into triviality: a preorder with this property is equivalent to the terminal category". Whoops, you're right! I had originally thought as much, then some silly mistaken reasoning had me thinking preorders with this property were actually quite common. $\endgroup$
    – Sridhar Ramesh
    Commented Nov 11, 2010 at 22:52

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It looks like such categories may be rather easy to construct. The following example should give the general idea: take the category of sets $V_\alpha$ of cardinality less than or equal to $\alpha$, for some infinite cardinal $\alpha$. The morphism classifier will be the set $C$ of cardinals up to and including $\alpha$, and the universal morphism should be $S \to C$ where the fiber over a cardinal $\beta$ is a set of cardinality $\beta$.

Given any function $f: Y \to X$, the classifying morphism $\chi_f: X \to C$ takes $x$ to the cardinal number of $f^{-1}(x)$.

Hopefully I haven't made any dumb mistakes...

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  • $\begingroup$ Nice. But does $C$ sit inside $V_\alpha$? I'm not objecting, I honestly don't know. I'm thinking of the case when $\alpha$ is inaccessible, and hence we have a model of ETCS, and having such a classifier seems rather strong. $\endgroup$
    – David Roberts ♦
    Commented Nov 12, 2010 at 0:44
  • $\begingroup$ I'm pretty sure it's all right: the collection of cardinals less than or equal to a given cardinal $\alpha$ is contained in the collection of ordinals less than the ordinal $\alpha + 1$, which in the von Neumann conception is the ordinal $\alpha + 1$, which has cardinality $\alpha$. Also, the collection $S$ has less than or equal to $\alpha$ elements in each fiber, ranging over points in the base (of which there are at most $\alpha$), so the $S$ is in bijection with a subset of $\alpha^2$ which has cardinality $\alpha$. $\endgroup$
    – Todd Trimble
    Commented Nov 12, 2010 at 2:03
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    $\begingroup$ Ah, I see a possible irritation with my notation: I think that usually people reserve $V_\alpha$ for a universe of sets of cardinality strictly less than $\alpha$ (so that we get a model of ETCS or ZFC for $\alpha$ strongly inaccessible), but here $V_\alpha$ consists of sets up to and including cardinality $\alpha$, which makes a big difference. $\endgroup$
    – Todd Trimble
    Commented Nov 12, 2010 at 2:07
  • $\begingroup$ Ah, of course. Both a pity and a joy that it can be so straightforward. The way a "set of all sets" (which is what C in some sense is, in that universe) often causes such problems had me worrying it would be trickier to pull off, but I suppose what this shows is that those problems only arise in the context of certain exponentials. $\endgroup$
    – Sridhar Ramesh
    Commented Nov 12, 2010 at 2:21
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    $\begingroup$ A) That last comment was in reply to the answer, not the comment above it. B) Actually, I think in general people usually use $V_{\alpha}$ for sets of rank less than $\alpha$. Still, you explained your use well enough to follow. $\endgroup$
    – Sridhar Ramesh
    Commented Nov 12, 2010 at 2:22
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I claim that a such category exist: Let $\mathcal{C}$ any category, build a full immersion $\mathcal{C}\subset \mathcal{C'} $ adding to $\mathcal{C}$ the new objects: $\coprod_f d_0(f), \ \coprod_f d_1(f) $ and the no-identity new arrows: $\coprod_f f: \coprod_f d_0(f) \to \coprod_f d_1(f)$ and for any $f: X\to Y$ the arrows $\epsilon^0_f: X\to \coprod_f d_0(f)$ (think as a f-coprojection of $X=d_0(f)$) and $\epsilon^1_f: X\to \coprod_f d_1(f)$. Then the $\mathcal{C'}$ arrows are the $\mathcal{C}$ arrow more the following:

$\coprod_ff\star \epsilon^0_g\star h $ , $ \star \epsilon^0_g\star h $, $ \epsilon^1_g\star h $ ($h\in \mathcal{C}\downarrow d_0(g) $) with obvious composition (here "$\star$" is a "free" composition, and composing morphisms of $\mathcal{C}$ by original compositionlaw " $\circ$ " whenever possible).

Then in $\mathcal{C'}$ we consider the congruence : $\coprod_ff\star \epsilon^0_f\star g \sim \epsilon^1_f\star (f\circ g) $

And let $\mathcal{C''}$ the quotient category, still we have a full immersion $\mathcal{C}\subset \mathcal{C''} $

And in $\mathcal{C''}$ we have the commutative diagram:

1] $$\require{AMScd} \begin{CD} X @>{\epsilon^0_f}>> \coprod_fd_0(f)\\ @V{f}VV @VV{\coprod_ff}V \\ Y @>{\epsilon^1_f}>> \coprod_fd_1(f) \end{CD}$$

i.e. $\coprod_ff\circ \epsilon^0_f = \epsilon^1_f \circ f$, but the follow isn't commutative (commutative only if $f=g=h$):

2] $$\require{AMScd} \begin{CD} X @>{\epsilon^0_g}>> \coprod_fd_0(f)\\ @V{h}VV @VV{\coprod_ff}V \\ Y @>{\epsilon^1_f}>> \coprod_fd_1(f) \end{CD}$$

i.e. $\coprod_ff\circ \epsilon^0_g \neq \epsilon^1_f \circ h$.

Then [1] is a Pullback, and $\coprod_f f: \coprod_f d_0(f) \to \coprod_f d_1(f)$ classifying any arrow of $\mathcal{C''}$.

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