Defining the notion of embedding of Grothendieck toposes as a quotient in the opposite category

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. ◼