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. $\endgroup$ – Harry Gindi Oct 14 '18 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 '18 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. $\endgroup$ – Alexander Campbell Oct 14 '18 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.) $\endgroup$ – Alexander Campbell Oct 14 '18 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 '18 at 15:02

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