Defining the notion of embedding of Grothendieck toposes as a quotient in the opposite category The category of locales is defined to be the opposite of the category of frames. That's why the notion of a sublocale is defined to be a quotient in the category of frames (subobjects are dual to quotients). (See here for a formal definition.)
Similarly, the category of Grothendieck toposes can be considered to be the opposite of the category of "algebraic" Grothendieck toposes. By the category of algebraic Grothendieck toposes I mean the category of Grothendieck toposes and functors preserving all colimits and finite limits. Reversing the order of these morphisms is equivalent to the usual notion of geometric morphism (by some adjoint functor theorems): adjunctions in which the left adjoint preserves finite limits.
Now, an embedding of Grothendieck toposes is a geometric morphism in which the right adjoint is fully faithful.
Question: Can this definition of embedding and subtopos be phrased as a quotient in the category of algebraic toposes? (Similarly as a sublocale can be defined as a quotient in the category of frames.)
Note that locales can be considered to be Grothendieck 0-toposes, so the analogy makes sense.
 A: In fact, the situation is exactly analogous.
We have the following general fact.
Proposition.
Consider an adjunction:
$$f^* \dashv f_* : \mathcal{S} \to \mathcal{T}$$
The following are equivalent:

*

*The left adjoint $f^* : \mathcal{T} \to \mathcal{S}$ is a localisation, i.e. for every category $\mathcal{X}$, precomposition by $f^*$ gives a fully faithful functor $[\mathcal{S}, \mathcal{X}] \to [\mathcal{T}, \mathcal{X}]$ identifying it with the full subcategory of functors $G : \mathcal{T} \to \mathcal{X}$ such that, for all morphisms $\phi$ in $\mathcal{T}$, $f^* \phi \text{ invertible} \implies G \phi \text{ invertible}$.


*The right adjoint $f_* : \mathcal{S} \to \mathcal{T}$ is fully faithful.
Localisations are one way of categorifying quotients.
I suspect the non-elementary nature of the definition of localisation is why the definition using the right adjoint is given first.
Incidentally, you can give an analogous definition for embeddings of locales: a frame homomorphism $f^* : M \to L$ is surjective if and only if, considered as a monotone map, its right adjoint $f_* : L \to M$ satisfies $f^* f_* = \textrm{id}_L$.
(Recall that a right adjoint functor is fully faithful if and only if the counit is an isomorphism!)

Proof of proposition.
First, suppose $f^* : \mathcal{T} \to \mathcal{S}$ is a localisation.
Then precomposition by $f^*$ gives a fully faithful functor $(f^*)^* :[\mathcal{S}, \mathcal{S}] \to [\mathcal{T}, \mathcal{S}]$.
But $f^* : \mathcal{T} \to \mathcal{S}$ has a right adjoint, so precomposition by $f^*$ has a left adjoint, namely precomposition by $f_* : \mathcal{S} \to \mathcal{T}$, $(f_*)^* : [\mathcal{T}, \mathcal{S}] \to [\mathcal{S}, \mathcal{S}]$.
(Yes, the right adjoint induces a left adjoint.
Please bear with me.)
The counit of the induced adjunction, $(f_*)^* (f^*)^* \Rightarrow \textrm{id}_{[\mathcal{S}, \mathcal{S}]}$, is horizontal precomposition by the counit of the original adjunction, $f^* f_* \Rightarrow \textrm{id}_\mathcal{S}$; but since $(f_*)^* : [\mathcal{T}, \mathcal{S}] \to [\mathcal{S}, \mathcal{S}]$ is fully faithful, $(f_*)^* (f^*)^* \Rightarrow \textrm{id}_{[\mathcal{S}, \mathcal{S}]}$ is a natural isomorphism, so its component at $\textrm{id}_\mathcal{S}$ is an isomorphism, i.e. $f^* f_* \Rightarrow \textrm{id}_\mathcal{S}$ is also a natural isomorphism.
Hence $f_* : \mathcal{S} \to \mathcal{T}$ is fully faithful.
Conversely, suppose $f_* : \mathcal{S} \to \mathcal{T}$ is fully faithful.
Then the counit $f^* f_* \Rightarrow \textrm{id}_\mathcal{S}$ is a natural isomorphism, so for any category $\mathcal{X}$, the induced counit $(f_*)^* (f^*)^* \Rightarrow \textrm{id}_{[\mathcal{S}, \mathcal{X}]}$ is also a natural isomorphism, so $(f^*)^* : [\mathcal{S}, \mathcal{X}] \to [\mathcal{T}, \mathcal{X}]$ is fully faithful.
It remains to be shown that the essential image is what we claimed.
Suppose $G : \mathcal{T} \to \mathcal{X}$ is a functor that inverts every morphism $f^* : \mathcal{T} \to \mathcal{S}$ inverts.
Consider the unit $\textrm{id}_\mathcal{T} \Rightarrow f_* f^*$.
Horizontally postcomposing by $f^*$ yields a natural isomorphism $f^* \Rightarrow f^* f_* f^*$, so horizontally postcomposing by $G$ yields a natural isomorphism $G \Rightarrow G f_* f^*$.
But $G f_* f^*$ is manifestly in the image of $(f^*)^* : [\mathcal{S}, \mathcal{X}] \to [\mathcal{T}, \mathcal{X}]$, so we are done. ◼
