16
$\begingroup$

I've never seen an example of a category with a subobject classifier which didn't embed nicely into a topos. Is there a good reason for this?

Question 1: Let $\mathcal C$ be a category with a subobject classifier $\Omega$ (and whatever finite limits this entails -- namely, a terminal object and pullbacks along monomorphisms). Does there exist a fully faithful functor $\mathcal C \to \mathcal E$, where $\mathcal E$ is an elementary topos, which preserves the subobject classifier and the aforementioned finite limits?

Question 2: Same as Question 1, but assuming that $\mathcal C$ has all finite limits, and requiring that $\mathcal C \to \mathcal E$ preserves them.

Question 3: Same as Question 2, but throwing in finite colimits as well.

Question 4: Now assume that $\mathcal C$ is locally presentable and has a subobject classifer $\Omega$. Does it follow that $\mathcal C$ is a (necessarily Grothendieck) topos?

Question 4 may be the most heavy-duty-sounding formulation, but it also gives me the most reason to think the answer might be "yes" -- after all, in order for a category $\mathcal C$ with finite limits and a subobject classifier to be a topos, it just needs to additionally be cartesian closed. And if $\mathcal C$ is locally presentable, then by the adjoint functor theorem, to verify this one just needs to check that the functors $X \times (-)$ preserve colimits. Plausibly, the subobject classifier might force this. As partial progress, I think I can show that in this case, coproducts are disjoint.

Improve this question
$\endgroup$
5
  • 8
    $\begingroup$ On Q4: The category of pointed sets is locally finitely presentable and has a subobject classifier. Yet it is not a topos. $\endgroup$
    – Ivan Di Liberti
    Commented Sep 17, 2020 at 20:35
  • 2
    $\begingroup$ @IvanDiLiberti Oh wow -- that's a great example! And the universal subobject is a map $1_+ \to 2_+$, where the domain $1_+$ is not the terminal object. I did not realize this was possible! $\endgroup$
    – Tim Campion
    Commented Sep 17, 2020 at 20:43
  • 3
    $\begingroup$ Hum, no the subobject classifier is $0_+ \to 1_+$ and $0_+$ is terminal. $\endgroup$
    – Simon Henry
    Commented Sep 17, 2020 at 21:07
  • 3
    $\begingroup$ Ah, right! In fact, this example seems to be general -- for any topos $\mathcal E$ with universal subobject $true: 1 \to \Omega$, the category $\mathcal E_\ast$ of pointed objects in $\mathcal E$ has a subobject classifier given by $(\Omega, true)$. But $\mathcal E_\ast$ is only a topos if it is the terminal category, which only happens if $\mathcal E$ is the trivial topos. $\endgroup$
    – Tim Campion
    Commented Sep 17, 2020 at 22:11
  • 3
    $\begingroup$ Not quite what you were asking, but there are pretoposes with subobject classifier (well-pointed, Boolean, and satisfying IAC, even) that are not cartesian closed, and very far from being (elementary) toposes. $\endgroup$
    – David Roberts ♦
    Commented Sep 17, 2020 at 22:54

1 Answer 1

Reset to default
17
$\begingroup$

Ivan's example in the comment actually proves that all the questions have negative answers.

As observed by Ivan, in the category of pointed set, there is a subobject classifier given by $\{*\} \to \{*,\bot \}$, where $*$ is the special point.

Indeed, a subobject of $X$, is just a subset of $X$ containing $*$ so it is classified by a unique map $X \to \{* = \top,\bot\}$ : the usual classifier of the map in Set.

Now, in a topos, you always have at least two maps from the terminal object to the sub-object classifier: the map $\top$ and the map $\bot$. If they are equal, the topos is degenerated. But in pointed set, there is only one map from $\{*\} \to \{*,\bot \}$, so there can't by a fully faithful functor to an elementary topos that preserve the subobject classifier and its universal subobject.

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ Maybe one comment. While the symmetric topos construction shows that finite limits can be formally added "lexely", providing a left adjoint to the forgetful functor Topoi $\to$ LP, this counterexample shows that cartesian closedness cannot be forced without destroying everything. $\endgroup$
    – Ivan Di Liberti
    Commented Sep 17, 2020 at 21:27
  • $\begingroup$ @Ivan "Lexely" or "laxly"? $\endgroup$
    – David Roberts ♦
    Commented Sep 17, 2020 at 22:51
  • 2
    $\begingroup$ @DavidRoberts I think Ivan did meant "lexely" as the lex colimits of arxiv.org/abs/1107.0778 $\endgroup$
    – Simon Henry
    Commented Sep 17, 2020 at 23:23
  • $\begingroup$ That is precisely what I meant. $\endgroup$
    – Ivan Di Liberti
    Commented Sep 18, 2020 at 6:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .