Properties of quotient categories.

Question

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:

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 and one can proove that $S$ is fully faithful.)
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 counterexamples for the second point of my question.
2. 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.
3. 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.

Answer 1

Here's a simple counterexample for (2).

Let $\mathcal{A}$ be the category of sequences of linear maps $U\stackrel{\alpha}{\to}V\stackrel{\beta}{\to}W$ between vector spaces over a field $k$ such that $\beta\alpha=0$, and let $\mathcal{C}$ be the full subcategory of objects of the form $0\to0\to W$, so that $\mathcal{A}/\mathcal{C}$ is equivalent to the category of linear maps $U\to V$, with the quotient functor sending $U\to V\to W$ to $U\to V$ and the section functor sending $U\to V$ to $U\to V\to0$.

Then $0\to k=k$ is injective in $\mathcal{A}$, but $T(0\to k=k)$ is $0\to k$, which is not injective in $\mathcal{A}/\mathcal{C}$.

For the projective case, you can use the same $\mathcal{A}$, but take $\mathcal{C}$ to be the category of objects $U\to0\to0$, so that $\mathcal{A}/\mathcal{C}$ is again equivalent to the category of linear maps, the quotient functor sends $U\to V\to W$ to $V\to W$, and the section functor sends $V\stackrel{\beta}{\to}W$ to $\ker(\beta)\to V\stackrel{\beta}{\to}W$.

Then $k=k\to0$ is projective in $\mathcal{A}$ but $T(k=k\to0$) is $k\to 0$, which is not projective in $\mathcal{A}/\mathcal{C}$.

Answer 2

A simple example of Grothendieck category without projectives is the category of torsion abelian groups. Another examples are the categories of sheaves.

About the other questions: If $\mathcal A$ is complete, then ${\mathcal A}/{\mathcal C}$ has to be also complete, since the fully faithful functor $S$ (which may be seen as an inclusion) has a left adjoint, hence it preserves limits. But, if $\mathcal A$ is complete then, an inclusion functor preserves limits iff the respective subcategory is closed under limits. See also Cocomplete but not complete abelian category

If $\mathcal A$ is Grothendieck, then it has enough injectives. The proof of this fact may be a consequence of Gabriel-Popescu theorem, as in B. Stenstrom, Rings of Quotients, Chap. X, Corollary 4.3, but may be shown directly by transfinite induction, as Grothendieck did in so called Tohoku paper:

Al. Grothendieck, "Sur quelques points d'algèbre homologique", The Tohoku Mathematical Journal. Second Series 9: 119–221

I think is a good question if the existence of injectives in $\mathcal A$ imply the same property for $\mathcal A/\mathcal C$.