# Exercise in Homological Algebra

I'm totally stuck with this problem that I found in an Algebra course. It is the following:

Let $$F:\mathcal{A} \to \mathcal{B}$$ be a left exact functor between two abelian cathegories. Let $$\mathcal{F}$$ be a family of objects of $$\mathcal{A}$$ with these two properties:

(i) For every object $$x$$ of $$\mathcal{A}$$ there is $$y\in \mathcal{F}$$ and a monomorphism $$\phi:x\to y.$$

(ii) for every short exact sequence $$0 \to a \to b \to c \to 0$$ with $$a,b \in \mathcal{F},$$ we have the properties $$c \in \mathcal{F}$$ and $$0 \to Fa \to Fb \to Fc \to 0$$ is exact.

Then for every complex bounded from below $$a^*$$ with $$a^n\in \mathcal{F}$$ for every $$n$$ we have a quasi-isomorphism $$F(a^*)\to RF(a^*).$$

$$RF(a)$$ is defined as follows: take the complex $$0\to a^0\to a^1\to ...$$, take an injective resolution $$0\to i^0\to i^1\to...$$ and consider the complex $$RF(a^*)=F(i^*)$$.

I showed that the functor $$RF$$ is well defined (up to homotopic equivalence) and sends distinguished triangles to distinguished triangles.

Using property (ii) I showed that $$F$$ preserves exactness for long exact sequences of the form $$0\to b^0\to b^1\to...$$ with $$b^n\in\mathcal{F}$$.

I still have no idea about how to use (i): maybe I may try to construct resolutions made of objects of $$\mathcal{F}$$? Would these be somehow related to injective resolutions? Sadly I didn't come out with anything. Thanks to everybody.

• The full subcategory of $\mathcal{A}$ generated by $\mathcal{F}$ is called an $F$-injective subcategory. You might be interested in [Definition 1.3.2 and lemma 1.3.3 here][1]. They cover all details (at least for the dual notion). [1]: numdam.org/issue/MSMF_1999_2_76__R3_0.pdf – Mathematician 42 Mar 5 at 7:00
• you should not use injective resolution, the properties i) and ii) give you a way of resolving any object via $\mathcal{F}$, in particular, it behaves like a injective resolution. (the idea of injective resolutions is just that every left exact functor is exact on them) So I would just construct a $\delta$-functor using $\mathcal{F}$ and then use univerality of $RF$. I could write down a proof, but this is just exhausting and fiddely. – Enkidu Mar 6 at 10:36