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An object $X$ of a cartesian closed category $\mathbf C$ is atomic if $({-})^X \colon \mathbf C \to \mathbf C$ has a right adjoint (hence is also internally tiny). Intuitively, atomic objects are "very small" (as the name suggests), and consequently there aren't usually many tiny objects in $\mathbf C$.

However, is this necessarily the case? More precisely, do there exist any non-posetal cartesian closed categories in which every object is atomic?

If there are no non-posetal examples, are there any nontrivial posetal examples? (The terminal category forms a trivial example of a posetal example.)

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  • $\begingroup$ Does it really imply tiny-ness ? It seems like this deduction uses that $*$ is tiny. $\endgroup$
    – Maxime Ramzi
    Commented Apr 27, 2022 at 11:15
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    $\begingroup$ I think we can show that there is no cocomplete category where every object is tiny. What follows is only an idea: when a cocomplete category K has a dense generator of tiny objects G, then K is Psh(G). Because of course, K is dense in itself, we would obtain that K is its own presheaf category. I think one can show that is is never the case. $\endgroup$
    – Ivan Di Liberti
    Commented Apr 27, 2022 at 11:24
  • $\begingroup$ @MaximeRamzi: thanks, I have clarified that I meant "internally tiny"; I overlooked the distinction between the two notions. $\endgroup$
    – varkor
    Commented Apr 27, 2022 at 12:12
  • $\begingroup$ @IvanDiLiberti : I think there are two issues with your sketch : 1- as I pointed out above, atomic need not imply tiny, 2- The equivalence "K has a dense generator of tiny objects G iff K = Psh(G)" holds only if G is small: you cannot define Psh(G) -> K by left Kan extension otherwise, because you did not assume K had large colimits (in other words, for your full sketch you need K to have K-sized colimits, i.e. K must be a poset) $\endgroup$
    – Maxime Ramzi
    Commented Apr 27, 2022 at 12:14
  • $\begingroup$ @MaximeRamzi do not forget the Wittgensteinian difference between to say and to show. $\endgroup$
    – Ivan Di Liberti
    Commented Apr 27, 2022 at 12:18

2 Answers 2

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Building on Maxime's answer -- if $C$ is cartesian closed and has an initial object $0$, and if $0$ is atomic (or even just tiny), then $C$ is the terminal category. For $1 = 0^0 = 0$ (the former equation holds in any cartesian closed category with an inital object $0$; the latter holds because $(-)^0$ preserves initial objects). That is, $C$ is pointed, and the only pointed cartesian closed category is the terminal one.

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    $\begingroup$ Thanks. I had spotted that having an initial object made the category pointed, but completely overlooked that this implied triviality. $\endgroup$
    – varkor
    Commented Apr 27, 2022 at 18:48
  • $\begingroup$ Exactly the same thing happened for me :D that's a nice observation ! $\endgroup$
    – Maxime Ramzi
    Commented Apr 28, 2022 at 7:16
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This is a partial answer : if $C$ has finite coproducts, then $C$ must be posetal. In fact, I only need biproducts of the form $X\coprod X$.

Indeed, because $C$ is cartesian closed, $X\times -$ commutes with these coproducts and in particular it is easy to check that $\underline{\hom}(* \coprod *, Y)\cong Y \times Y$, naturally in $Y$ ($\underline\hom$ denotes the internal hom).

In particular, if $*\coprod *$ is atomic, the canonical morphism $(Y_0\times Y_0)\coprod (Y_1\times Y_1)\to (Y_0\times Y_0)\coprod (Y_0\times Y_1)\coprod (Y_1\times Y_0)\coprod (Y_1\times Y_1)$ is an isomorphim (here I'm denoting by $Y_0$ or $Y_1$ the same object $Y$, it's simply to indicate the "position", i.e. what the morphism is).

This morphism is of the form $X\overset{in_1}\to X\coprod X$, and the claim is that this is an isomorphism. This implies that any two morphisms $X\to Z$ must be equal: $\hom(X,Z)\times \hom(X,Z)\cong \hom(X,Z)$.

Here $X$ is $Y\times Y$, for an arbitrary $Y$. Any of the two projections $Y\times Y\to Y$ is split, so it follows that any two morphisms out of $Y$ must be equal. $Y$ was arbitrary, so $C$ is posetal.

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    $\begingroup$ Thank you, this is already very strong evidence that atomocity of every object is an extremely restrictive assumption. In other words, every tiny object must be "superinitial" (assuming enough coproducts). $\endgroup$
    – varkor
    Commented Apr 27, 2022 at 13:16

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