MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{C}$ be abelian categories and $F: \mathcal{A} \to \mathcal{C}$ and $G: \mathcal{B} \to \mathcal{C}$ be functors between them. It is then possible to define the pullback category $\mathcal{D} := \mathcal{A}\times_\mathcal{C} \mathcal{B}$, which is abelian again. One way to define $\mathcal{D}$ concretely is the following:

  • The objects of $\mathcal{D}$ consist of pairs $(A,B)$, where $A$ is an object of $\mathcal{A}$ and $\mathcal{B}$ is an object of $B$, together with an isomorphism $f: FA \to GB$.
  • Morphisms between $(A,B,f)$ and $(A', B', f')$ consist of two morphisms $A\to A'$ and $B\to B'$ such that the corresponding square diagram in $\mathcal{C}$ commutes.

Such situations sometimes occur in algebraic geometry, where the relevant abelian categories are categories of quasi-coherent modules.

By definition, we get an exact sequence \begin{eqnarray*}0\to Hom_{\mathcal{D}}((A,B,f), (A',B',f')) \to Hom_{\mathcal{A}}(A,A') \oplus Hom_{\mathcal{B}}(B,B') \to Hom_{\mathcal{C}}(FA,GB')\end{eqnarray*}

Now suppose that all occuring abelian categories have, say, enough injectives. Then my question is the following:

Can we define boundary morphisms \begin{eqnarray*}Ext^i_{\mathcal{C}}(FA, > GB') \to Ext^{i+1}_{\mathcal{D}}((A, > B, f), (A',B',f'))\end{eqnarray*} such that the emerging sequence of Ext groups is long exact?

Probably one needs at least some additional assumption for this to hold, such as exactness of $F$ and $G$ or the existence of some adjoints.

This is an attempt to control the cohomological dimension of $\mathcal{D}$ in terms of the cohomological dimension of $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{C}$ - other suggestions to this enterprise are also welcome.

share|cite|improve this question
It looks a bit like you want to "glue" stuff together. If so, then this paper by Polishchuk (and the references thereein) seems relevant: In the introduction it is stated that the gluing need not necessarily be of finite cohomological dimension even if the original categories are of finite cohomological dimension (I haven't looked very closely, though). – Theo Buehler Jul 22 '11 at 18:07
Lennart, do you assume that $F,G$ are exact? Otherwise $A \times_C B$ has no chance to be abelian. – Martin Brandenburg Jul 22 '11 at 20:10
Also, have you checked the example that $F,G$ are restriction functors from quasi-coherent modules along open subsets? – Martin Brandenburg Jul 22 '11 at 20:13
@Theo: What they do, seems to be slightly different. @Martin: Yes, F and G should be exact. For the case of quasi-coherent modules on an open covering it works. – Lennart Meier Mar 17 '12 at 21:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.