Let $\mathcal{C}$ be a category and $\mathcal{W} \subseteq \text{Arr}(\mathcal{C})$ a set (or class) of arrows.


There are (at least) two notions of localization of $\mathcal{C}$ with respect to $\mathcal{W}$, a strict and a weak one.
In the strict one, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is a localization iff $\lambda$ inverts all arrows in $\mathcal{W}$ and if for every functor $G: \mathcal{C} \rightarrow \mathcal{E}$ with the same property there exists a *unique* functor $G':\mathcal{D} \rightarrow \mathcal{E}$ with
$$ G = G'\circ{}\lambda $$
It is not hard to see, that such a strict localization functor $\lambda$ must actually be surjective on objects.
In the weak version of localization, one merely asks for the existence of a functor $G'$ together with an isomorphism
$$ \eta: G \stackrel{\sim}{\rightarrow} G'\circ{}\lambda$$
such that for every other such pair $(G'',\eta')$ there exists a unique isomorphism $\kappa:G' \stackrel{\sim}{\rightarrow} G''$ with
$$ (\kappa \circ{} \lambda) \eta = \eta' $$
Now, if such a weak localization functor would always be essentially surjective, then I (think I) can prove that the notions of strict and weak localization are essentially equivalent, in the sense that the existence of one implies that of the other.


So, are all weak localization functors essentially surjective?



Are localization functors always essentially surjective?
For any weak version you have a stong one where $\lambda$ is the identity on objects. Moreover, the target of a (weak) localization functor is essentially unique (up to equivalence). Essential surjectivity is invariant by equivalences, hence the answer to your question is positive.


