# Are there right-deformations for abelian sheaves?

A sufficient condition for the existence of a point-set derived functor is the existence of a deformation of the corresponding functor. For modules, such a deformation always exists (see section 2.3). Generally, however, even projective resolutions are not guaranteed to be functorial.

Since abelian sheaves form a Grothendieck category, and hence have functorial injective embeddings, I was wondering whether this is enough to actually give a deformation for each functor on the category of (positively graded) chain complexes of abelian sheaves. If so, how does one construct it from functorial embeddings?

Constructing functorial injective resolutions $\mathsf{Ab}(X)\longrightarrow \mathsf{Ch}^{\geq 0}_\bullet (\mathsf{Ab}(X))$ seems easy using splicing and the fact images and kernels are also functors. I don't know how to construct a deformation $\mathsf{Ch}^{\geq 0}_\bullet (\mathsf{Ab}(X))\longrightarrow \mathsf{Ch}^{\geq 0}_\bullet (\mathsf{Ab}(X))$ though, not to mention a deformation for a given functor...

So:

Does each functor between categories of abelian sheaves have a deformation, and how can one construct it from the functorial injective embedding?

Added Later: See Definitions 2.2.1,2.2.4 here.

• I don't understand what you mean by "deformation" here. Can you clarify? – Qiaochu Yuan Oct 10 '15 at 21:23
• @QiaochuYuan I added the general definition. Here, we take the weak equivalences of $\mathsf{Ch}^{\geq 0}_\bullet (\mathsf{Ab}(X))$ to be quasi-isomorphisms. – Arrow Oct 10 '15 at 21:36
• You are missing an important point, which is that deformations are supposed to be chosen for the specific functor you want to derive. Notice that the identity is trivially a deformation. – Zhen Lin Oct 10 '15 at 21:47
• @ZhenLin horrible mistake on my part. Thanks for pointing it out. Does my edited question make sense? – Arrow Oct 10 '15 at 22:01
• You haven't really changed the question – as it stands, it still admits a trivial answer. I can guess what you're really trying to ask, but it would much better if you could work it out yourself. – Zhen Lin Oct 10 '15 at 22:29