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?



When is a reflective subcategory of a topos a topos?
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.


