Let $\mathfrak{C}$ be a category isomorphic (or equivalent) to an elementary topos $\mathfrak{D}$. Is it true that also $\mathfrak{C}$ is an elementary topos? Similarly, if $\mathfrak{D}$ is a Grothendieck topos, is also $\mathfrak{C}$ a Grothendieck topos?

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    $\begingroup$ Of course, since the definition of a (Grothendieck) topos is stated in the language of category theory. And also, nobody would study a property of categories which is not invariant under equivalence ("categories which have exactly 42 objects"); this is not a proof of course, but shows you "in advance" that the answer must be "Yes" to your question. $\endgroup$ Commented Apr 28, 2021 at 11:58
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    $\begingroup$ @MartinBrandenburg's comment is the true answer. The two currently given answers are worthwhile in that they spell out the details, but you shouldn't be mislead into thinking that the statement would require such detailed proofs. The metatheorem Martin reveals that the analogous question with "elementary topos" replaced by any other category-theoretic notion has the same positive answer. (To Martin and other people interested in logic: Do we have a reference for a metatheorem which is directly applicable for the Grothendieck case? Being a Grothendieck topos is not an elementary statement.) $\endgroup$ Commented Apr 28, 2021 at 19:00
  • $\begingroup$ @IngoBlechschmidt What do you mean by "Being a Grothendieck topos is not an elementary statement"? :-) $\endgroup$ Commented Apr 28, 2021 at 21:28
  • $\begingroup$ @MartinBrandenburg I mean that it can not be formulated purely in the formal language of category theory (I mean the language which has one sort/type for the objects, one sort (or a dependent type) for the morphisms, etc.). Of course this language is very restrictive and lots of generalizations are useful :-) $\endgroup$ Commented Apr 29, 2021 at 10:41
  • $\begingroup$ Hmm, if you take the definition to be that a Grothendieck topos is a category equal to the category of sheaves on a Grothendieck site, I'm having trouble seeing how that's necessarily preserved under equivalences. Of course, if you tweak the definiition to replace "equal to" with "equivalent to" then that makes being preserved under equivalence to be trivial. $\endgroup$ Commented Apr 29, 2021 at 14:22

2 Answers 2


Alessandro has given a concrete solution for elementary toposes; the answer for Grothendieck toposes is also positive. I've also included a more abstract approach to proving the statement for elementary toposes.

An elementary topos is a cartesian-closed finitely complete category with a subobject classifier.

Recall that every equivalence can be improved to an adjoint equivalence . Hence, equivalences preserve and reflect finite limits. Since adjunctions compose, and cartesian-closure is defined in terms of $({-}) \times A$ having a right adjoint for all $A \in \mathscr C$, a category equivalent to a cartesian-closed category is also cartesian-closed (recalling that equivalences preserve finite products). Finally, note that having a subobject classifier is a representability condition, i.e. $\mathrm{Sub}_{\mathscr C}({-}) \cong \mathscr C({-}, \Omega)$. We can therefore compose with the equivalence to show that the subobjects of $\mathscr D$ are similarly representable. Hence, a category equivalent to an elementary topos is an elementary topos.

A Grothendieck topos is a left exact reflective subcategory of a presheaf category. Hence, since equivalences can be improved to adjoint equivalences, and are fully faithful, we can compose the equivalence with the reflection of $\mathscr D$ into a presheaf category.

lex reflection

Since equivalences preserve limits, $\mathscr C$ is therefore also a Grothendieck topos.

  • $\begingroup$ The link you gave on the Grothendieck topos says "... up to equivalence every Grothendieck topos arises this way: ..." I don't think the Grothendieck toposes are invariant with respect to equivalence: for example, in $\mathbf{SmallSet}$ if you take things literally in ZFC, and a functor is encoded as an ordered pair of the object function and the morphism function, then not every object in $\mathbf{SmallSet}$ is an ordered pair and therefore those objects cannot be presheaves on any category. $\endgroup$ Commented Apr 29, 2021 at 14:31
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    $\begingroup$ @DanielSchepler: I think one can reasonably take lex reflectivity in a presheaf category to be the definition of a Grothendieck topos. For an alternative proof, you can use that a Grothendieck topos is a locally presentable elementary topos, and show that local presentability is preserved by equivalence, which follows for similar reasons to preservation and reflection of colimits by equivalences. $\endgroup$
    – varkor
    Commented Apr 29, 2021 at 15:14

Well, looking at the definition from nLab, a topos $\mathscr T$ is a category that

  • Has finite limits
  • Is cartesian closed
  • Has a subobject classifier

Clearly, a category $C$ equivalent to a topos $\mathscr T$ has finite limits as well.

To prove that $C$ is cartesian closed, consider a pair of object $c,d\in C$ and let $F:C\rightarrow\mathscr T$ be an equivalence, then $$\mathbf{Hom}_C(e\times c,d)\simeq\mathbf{Hom}_{\mathscr T}(F(e\times c),F(d))\simeq\mathbf{Hom}_\mathscr{T}(F(e)\times F(c),F(d))=\star$$where I used that $F$, being an equivalence, preserves limits (and products in particular). Now, let $F(d)^{F(c)}$ be the exponential object of the pair $F(d),F(c)$, so that for every $z\in\mathscr T$ $$\mathbf{Hom}_\mathscr{T}(z\times F(c),F(d))\simeq\mathbf{Hom}_\mathscr{T}(z,F(d)^{F(c)})$$ so, continuing the previous equivalences $$\star\simeq\mathbf{Hom}_\mathscr{T}(F(e),F(d)^{F(c)})\simeq\mathbf{Hom}_C(e,G(F(d)^{F(c)}))$$where $G:\mathscr T\rightarrow C$ is the "inverse" equivalence of $F$. So $C$ is cartesian closed and an exponential object for $c,d$ is $G(F(d)^{F(c)})$.

Finally, to prove that $C$ has a subobject classifier, you can check this question Equivalence of categories preserves subobject classifiers.


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