Is it possible for a nontrivial category to have a slice classifier?

Question

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?

Answer 1

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...

Answer 2

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 $\mathcal{C''}$.