There are standard notions of "surjections" and "embeddings" of toposes. However, not every surjection is an epimorphism, and not every regular monomorphism is an embedding. (EDIT: as Alexander Campbell points out in the comments, in this higher context, regular mono does not imply mono. So perhaps embeddings are not as strange as I make them out to be here. I still find it strange that most surjections are not epimorphisms, though.)

Let $f: \mathcal Y \to \mathcal X$ be a geometric morphism. Recall that $f$ is said to be

  • surjective if $f^\ast$ is conservative (or equivalently, $f^\ast$ is comonadic.)

  • an embedding if $f_\ast$ is fully faithful (equivalently, $f^\ast$ is a localization).

I'd say the correct notion of monomorphism / epimorphism is the $(\infty,1)$-categorical one: $f$ is a monomorphism iff the canonical square $f \circ 1 = f \circ 1$ is a pullback, and dually for epimorphisms. Since $(\infty,1)$-colimits of topoi are computed by taking $(\infty,1)$-limits of the inverse image functors between the underlying categories, $f$ is an epimorphism iff $f^\ast$ is a monomorphism of $(\infty,1)$-categories. That is,

  • $f$ is an epimorphism iff $f^\ast$ is a replete subcategory inclusion, i.e. $f^\ast$ reflects the property of being isomorphic and is an inclusion of path components on hom-spaces.

(Note that the coalgebras for any accessible left exact comonad on $\mathcal Y$ is an $\infty$-topos $\mathcal X$ which admits a canonical surjection from $\mathcal Y$ which will typically not be an epimorphism.)

As for monomorphisms, clearly if $f$ is an embedding, then it is a monomorphism. But not even every "regular monomorphism" is an embedding (EDIT: which is not to say that not every monomorphism is an embedding -- see Alexander Campbell's comment below!). For example, if $F,G: C \to D$ are functors, then the $(\infty,1)$-equalizer of the induced geometric morphisms $Psh(C) \to Psh(D)$ is presheaves on the iso-inserter of $F$ and $G$. The canonical map $Psh(IsoIns(F,G)) \to Psh(C)$ typically fails to be an embedding. Anyway, this leaves me with the question:

Question: What are the monomorphisms of topoi, or of $\infty$-topoi?

I expect this may be complicated, given how complicated monomorphisms of affine schemes are (a category which behaves in some ways similarly to the category of toposes).

Note that because every embedding is a monomorphism, by the surjection / embedding factorization system it suffices to determine which surjections are monomorphisms.

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    $\begingroup$ I'm wondering if the way to make sense of this is to consider the $\infty,2$-category. If you do that, instead of epi-mono, eso-ff becomes the choice of factorization, and ff is equivalent to embedding. $\endgroup$ Oct 14, 2018 at 3:44
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    $\begingroup$ Even if a more 2-categorical factorization system is more convenient for some purposes, it's still sometimes actually useful to know that a particular morphism is a monomorphism. $\endgroup$
    – Tim Campion
    Oct 14, 2018 at 4:07
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    $\begingroup$ Note that equalizers need not be monomorphisms in higher contexts. For example, if G is a group seen as a one-object groupoid, then the inserter of the pair of morphisms G --> G consisting of the identity morphism and the constant morphism at the unit element is a contractible; but a morphism of groupoids is a monomorphism in the higher sense iff it is fully faithful. $\endgroup$ Oct 14, 2018 at 5:03
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    $\begingroup$ Your question arises in the paper 'Van Kampen theorems for toposes' of Bunge and Lack. In the paragraph before Theorem 5.1, they say that they do not know of a pseudomonic geometric morphism which is not an inclusion (i.e. an embedding). (Note that the term "pseudomonic" is used in the 2-categorical literature for monomorphisms in the higher sense.) $\endgroup$ Oct 14, 2018 at 5:09
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    $\begingroup$ @Alexander Thanks! The lack of examples surprises me. For example, if $f_\ast$ is a pseudomonomorphism of categories, then $f$ will be a pseudomonomorphism of toposes, which at least gives somewhere to look for examples. Maybe even pseudomonomorphisms of this type are hard to find... $\endgroup$
    – Tim Campion
    Oct 14, 2018 at 15:02

1 Answer 1


Edit: My original answer contained a big mistake, that I can't fix. A long time I had thought ago about monomorphisms of locales, and I wrongly convince myself that everything would generalizes to toposes, but I now realize things are more complicated than this... All my apologies about this.

I still can use what I said before to give an example of monomorphisms that are not embeddings, which is what I will explain now.

Given a locale $L$, there is an other locale $DL$, called the dissolution of $L$, endowed with a geometric morphism $ DL \rightarrow L$, which is universal for geometric morphism $p:K \rightarrow L$ such that for each open subspace $u \subset L$ (I.e. elements $u \in \mathcal{O}(L)$ of the frame corresponding to $L$), $p^* u$ is a complemented open subspace in $K$. $DL$ is explicitly constructed as the frames of nuclei of $L$, see for example Sketches of an elephant section C.1 for this construction.

Claim: For any non-boolean locale $L$, the geometric morphism $Sh(DL) \rightarrow Sh(L)$ is a monomorphisms of topos (or $\infty$-topos) that is not an embedding.

Indeed, $DL \rightarrow L$ is always a surjection, so if it is an embeddings it is an isomorphism, which only happen when $L$ is boolean.

Moreover, the map $DL \rightarrow L$ is also clearly a monomorphism in the $1$-category of locale: maps to $DL$ are a subset of maps to $L$ (those that send every open to a complemented element)

But as the functor $L \mapsto Sh(L)$ from the category of locales to the category of topos/$\infty$-topos is a right adjoint, it preserves finite limits and monomorphism. So $Sh(DL) \rightarrow Sh(L)$ is a monomorphism in the category of topos/ $\infty$-topos as well.

  • $\begingroup$ Given a topos $T$ and a boolean locale $B$, what does $T(B)$ mean?! $\endgroup$ Sep 20, 2019 at 22:20
  • $\begingroup$ And when you say $Hom(B,T)$ you mean $\mathsf{Topoi}(\mathsf{Sh}(B), T)$, right? $\endgroup$ Sep 20, 2019 at 22:32
  • $\begingroup$ Cool, thanks! Does $Hom(B,T)$ denote the category of geometric morphisms or the groupoid? If it means the latter, I think this is equivalent to the definition I was using -- i.e. that the groupoid inclusion $Hom(X,E) \to Hom(X,T)$ is an $\infty$-categorical monomorphism of 1-groupoids -- i.e. fully faithful. But if you mean the former, then your condition is stronger than mine -- by the tensoring of $Topoi$ over $Cat$, my definition means that $Hom(X,E) \to Hom(X,T)$ is equivalent to a full-on-isomorphisms (but not necessarily full) subcategory inclusion if $Hom$ denotes a category. $\endgroup$
    – Tim Campion
    Sep 21, 2019 at 15:58
  • $\begingroup$ Note that the dissolution map DL --> L, as a geometric morphism, is only a pseudomonic, but not a fully faithful morphism in general. For example, the map from discrete two-point-space to Sierpinski space does not reflect the specialization order. The simplest fully faithful morphism example I can find is L = (natural numbers + infinity), where neighborhoods of infinity are cofinite. Since this L already has a basis of complemented opens, there are no transformations between geometric morphisms to Sh(L) except for isomophisms. $\endgroup$ Nov 4, 2020 at 13:04
  • $\begingroup$ True, but the definition of "monomorphism" in the question (in terms of pullback square) corresponds to what I think you call pseudo-monic (more precisely: fully faithfull on invertible 2-cells). $\endgroup$ Nov 4, 2020 at 13:40

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