I asked this on [math.stackexchange.com][1], but didn't get any answer.


Let $\mathcal{A}$ be an abelian category and $\mathcal{C}$ a localizing subcategory in the sense of [Gabriel][2]. (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:

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

I know that:

 1. If $\mathcal{A}$ is cocomplete then so is $\mathcal{A}/\mathcal{C}$. ($T$ is a left adjoint)
 2. If $\{U_i\}$ is a set of generators then so is $\{T(U_i)\}$.
 3. 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.)
 4. 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.)
 5. If $\mathcal{A}$ is complete with respect to finite limits then so is $\mathcal{A}/\mathcal{C}$. ($T$ is exact.)

Edit:

 1. Jeremy Rickard gave a nice counterexample for (2).

  [1]: http://math.stackexchange.com/questions/402042/properties-of-quotient-categories
  [2]: http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1962__90_/BSMF_1962__90__323_0/BSMF_1962__90__323_0.pdf