Suppose $\mathcal{E}$ is a topos and $\mathcal{F}\subseteq \mathcal{E}$ is a reflective subcategory with reflector $L$, say. Under what conditions is $\mathcal{F}$ a topos?

A well-known sufficient condition for this is that $L$ be left exact. But this is certainly not necessary. For instance, let $f\colon C\to D$ be a functor such that $f^*\colon Set^D \to Set^C$ is fully faithful. A sufficient condition for this is given in C3.3.8 of Sketches of an Elephant — for every $d\in D$ the category of $c\in C$ with $d$ exhibited as a retract of $f(c)$ must be connected, and every morphism of $C$ must be a retract of the $f$-image of some morphism of $C$. These conditions do not imply that $\mathrm{Lan}_f$ is left exact, but nevertheless they allow us to identify $Set^D$ with a reflective subcategory of $Set^C$, and of course both are toposes.

Is there any general sufficient condition for a reflective subcategory of a topos to be a topos which includes this case?

  • $\begingroup$ from "Topos Theory" (p.Johnstone), Exercise1 p. 43, a general condiction is that the (reflexive) subcategory has power objects, and is ereditary for subobjects of (each) power object. $\endgroup$ – Buschi Sergio Dec 11 '11 at 9:57
  • $\begingroup$ @Buschi Sergio - sure, but that condition is almost identical to saying "the subcategory is a topos". I'd like some condition that is easier to verify! $\endgroup$ – Mike Shulman Dec 11 '11 at 23:06
  • $\begingroup$ What is the role of the reflector? $\endgroup$ – David Carchedi Dec 12 '11 at 14:31
  • $\begingroup$ I find this Street article, may be you just know it "Notions of topos" R. Street journals.cambridge.org/… $\endgroup$ – Buschi Sergio Jun 3 '12 at 20:30
  • $\begingroup$ @Buschi Sergio - Yes, I know it. I don't think it answers the question. $\endgroup$ – Mike Shulman Jun 5 '12 at 3:16

Let me pay no attention to size issues:

Denote the adjunction by $R$ right adjoint to $L$. Equip $\mathcal{E}$ with the canonical topology $J$ (so generated by jointly surjective epimorphisms), so that we have $Sh_J(\mathcal{E}) \simeq \mathcal{E}.$ Denote the induced sheafication functor $a:Set^{\mathcal{E}^{op}} \to \mathcal{E}.$ Now consider the Yoneda embedding $y:\mathcal{F} \hookrightarrow Set^{\mathcal{F}^{op}}.$ Since $\mathcal{F}$ is reflective in $\mathcal{E}$, it is cocomplete, so we can left-Kan extend the identify functor of $\mathcal{F}$ along Yoneda, to get a functor $a_\mathcal{F}:Set^{\mathcal{F}^{op}} \to \mathcal{F}$ which, by construction is left-adjoint to the Yoneda embedding.. EDIT: Since $\mathcal{E}$ is a topos, and therefore total, so is the reflective subcategory $\mathcal{F}.$ So we get to get a functor $a_\mathcal{F}:Set^{\mathcal{F}^{op}} \to \mathcal{F}$ which is left-adjoint to this Yoneda embedding. However, it is also canonically equivalent to $L \circ a \circ R_{!}$ since this composite is colimit preserving and along representables is the identity. So, it follows that $\mathcal{F}$ is a (Grothendieck) topos if and only if the composite $L \circ a \circ R_{!}$ is left-exact.

Note: By one of the comments I made below, $a \circ R_!$ is left-exact, so $\mathcal{F}$ is a topos if and only if $L$ preserves those finite limits of the form $\rho:\Delta_{aR_!(C)} \Rightarrow a \circ R_! \circ G,$ with $G:D \to Set^{\mathcal{F}^{op}}$ a finite diagram.

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    $\begingroup$ Interesting approach! I think your ignoring of size issues is in danger of being wrong when you construct $a_{\mathcal{F}}$ by "cocompleteness" of $\mathcal{F}$, since $\mathcal{F}$ will almost never have colimits of the same size as itself. However, I think you're in luck: any reflective subcategory of a total category (nlab.mathforge.org/nlab/show/total+category) is total, and that suffices to show the existence of $a_{\mathcal{F}}$. Why is $L \circ a \circ R_!$ the identity on representables? $\endgroup$ – Mike Shulman Dec 12 '11 at 18:10
  • $\begingroup$ Ah, good point. I guess I should be careful about sizes. Anyhow, $R_!$ is the left-Kan extension along Yoneda of the functor $\mathcal{F} \to Set^{\mathcal{E}^{op}},$ which sends $X \in \mathcal{F}$ to $Hom_{\mathcal{E}}(blank,X),$ which is representable. And, $a$ is the left-Kan extension along Yoneda of the identity functor, so it follows that for $X \in \mathcal{F},$ $a\circ R_!{X}=X$ which is secretly in $\mathcal{F},$ so the reflector does nothing, yielding $L \circ a \circ R_!(X)=X.$ $\endgroup$ – David Carchedi Dec 12 '11 at 18:27
  • $\begingroup$ Just an additional cute remark: Notice that $\mathcal{F}$ has finite limits. Now, $R_!$ is the left-Kan extension along Yoneda of $y \circ R$, which is left-exact, hence flat, because of the finite limits in $\mathcal{F}$, hence $R_!$ is left-exact. So is $a$. In particular, this tells us something we already knew, that if $L$ is left-exact, then $\mathcal{F}$ is a topos. $\endgroup$ – David Carchedi Dec 12 '11 at 19:12
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    $\begingroup$ That's actually quite cute! Note that it suffices to restrict even further to ask only that $L$ preserve pullbacks in the image of $a \circ R_!$, since any reflector preserves the terminal object. I wonder whether there is a way to refine this into something more manageable that doesn't refer to arbitrary pullbacks of presheaves on the large category $\mathcal{F}$. $\endgroup$ – Mike Shulman Dec 13 '11 at 6:38
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    $\begingroup$ Wait a minute, I have another question. Why does $R_!$ map $Set^{\mathcal{F}^{op}}$ into $Set^{\mathcal{E}^{op}}$ (as opposed to $SET^{\mathcal{E}^{op}}$)? Remember $\mathcal{E}$ and $\mathcal{F}$ are large categories, while $Set$ is the category of small sets. $\endgroup$ – Mike Shulman Jan 3 '12 at 18:18

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