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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.

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  • $\begingroup$ 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 $\endgroup$ – Mathematician 42 Mar 5 at 7:00
  • $\begingroup$ 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. $\endgroup$ – Enkidu Mar 6 at 10:36

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