Left exact functor $F$ preserves quasi-isomorphism between $F$-acyclics

Question

In this math overflow page, the poster gives a proof of the statement "an additive left exact functor $F$ preserves quasi-isomorphisms between $F$-acyclic objects." I'm having trouble understanding the final step of his proof, and I was wondering if someone here can enlighten me.

Suppose we have an additive left exact functor $F:A\rightarrow B$ between abelian categories. Then let $f:K\rightarrow L$ be a map of chain complexes where K and L consist of objecs that are $F$-acyclic. We have a short exact sequence

$$0\rightarrow K\rightarrow \operatorname{Cyl}(f)\rightarrow \operatorname{Cone}(f)\rightarrow 0.$$

Recall that the mapping cone of $f$ is acyclic presicely when $f$ is a quasi-isomorphism, and that the mapping cylinder of $f$ is quasi-isomorphic with $L$. Since additive functors preserve mapping cones and mapping cylinders (as they preserve direct sums), and $K$ is $F$-acyclic, we obtain the short exact sequence

$$0\rightarrow F(K)\rightarrow\operatorname{Cyl}(F(f))\rightarrow \operatorname{Cone}(F(f))\rightarrow 0.$$

According to the poster of the aforementioned post, the result should now follow. However, I don't see how it does. Most likely we are to prove that $\operatorname{Cone}(F(f))$ has trivial homology groups. Can someone help me with this?

Answer 1

It's not true, without boundedness conditions, that a left exact functor always preserves quasi-isomorphisms between complexes of $F$-acyclic objects.

As alluded to in the question, a chain map is a quasi-isomorphism if and only if its mapping cone is acyclic, and so the question is equivalent to the question of whether $F$ preserves acyclic complexes.

Let $R=\mathbb{Z}/4\mathbb{Z}$, and let $$X=\cdots\to\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times2}\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times2}\mathbb{Z}/4\mathbb{Z}\to\cdots,$$ an acyclic complex of $R$-modules whose terms are injective, and therefore $F$-acyclic for every left exact functor $F$. Then $F=\operatorname{Hom}_R(\mathbb{Z}/2\mathbb{Z},-)$ is a left exact functor, but $FX$ is $$X=\cdots\to\mathbb{Z}/2\mathbb{Z}\xrightarrow{0}\mathbb{Z}/2\mathbb{Z}\xrightarrow{0}\mathbb{Z}/2\mathbb{Z}\to\cdots,$$ which is not acyclic.

I suspect that in the post linked to, the poster forgot to include the assumption that the complexes are bounded to the left. With this extra condition on $X$, then it is true that if $X$ is an acyclic complex of $F$-acyclic objects, then $FX$ is acyclic, by standard facts about $F$-injective resolutions (at least if we assume that the category $A$ has enough injectives). Without loss of generality, $X$ is concentrated in nonpositive degrees, and so is an $F$-injective resolution of the zero object $0$, and so $FX$ is acyclic.