I asked this on math.stackexchange.com, but didn't get any answer.

Let $\mathcal{A}$ be an abelian category and $\mathcal{C}$ a localizing subcategory in the sense of Gabriel. (A Serre/thick/dense subcategory, such that the quotient functor $T\colon \mathcal{A}\rightarrow\mathcal{A}/\mathcal{C}$ admits a right adjoint, the so called "section functor".) Then one can form the quotient category $\mathcal{A}/\mathcal{C}$.

Which properties inherits $\mathcal{A}/\mathcal{C}$ from $\mathcal{A}$? To be more precise:

- If $\mathcal{A}$ has enough injectives (resp. projectives), does $\mathcal{A}/\mathcal{C}$ too? If not, under which conditions?
- If $A\in \mathcal{A}$ is injective (resp. projective), is it $T(A)$, too? If not, under which conditions?
- If $A\in \mathcal{A}$ is a cogenerator, is it $T(A)$ too? If not, under which conditions?
- If $\mathcal{A}$ is complete, is it $\mathcal{A}/\mathcal{C}$ too? If not, under which conditions?

I know that:

- If $\mathcal{A}$ is cocomplete then so is $\mathcal{A}/\mathcal{C}$. ($T$ is a left adjoint and one can proove that $S$ is fully faithful.)
- If $\{U_i\}$ is a set of generators then so is $\{T(U_i)\}$.
- If $\mathcal{A}$ is AB5 then so is $\mathcal{A}/\mathcal{C}$. ($T$ commutes with filtered limits and one can prove that taking filtered limits is exact.)
- The second and the third point implies: If $\mathcal{A}$ is Grothendieck then so is $\mathcal{A}/\mathcal{C}$. (Gabriel–Popesco theorem even says that every Grothendieck category has this shape and one can choose $\mathcal{A}$ as a category of modules.)
- If $\mathcal{A}$ is complete with respect to finite limits then so is $\mathcal{A}/\mathcal{C}$. ($T$ is exact.)

Edit:

- Jeremy Rickard gave counterexamples for the second point of my question.
- I answered the fourth point by myself: Let $F\colon \mathcal{D}\rightarrow \mathcal{E}$ be a functor with a fully faithful left adjoint $G$. Completeness of $\mathcal{D}$ implies completeness of $\mathcal{E}$ in general and this applies to the situation of quotient categories.
- Jeremy Rickard also answered the first question for projectives: There are Grothendieck categories with not enough projectives. They form a counterexample since they are quotients of a category of modules (Gabriel-Popescu) and every category of modules has enough projectives.