Let $F : \mathcal{A}\rightarrow\mathcal{B}$ be an additive functor of abelian categories, such that $F$ has cohomological dimension $\le n$. Suppose $\mathcal{A}$ has enough injectives. Let $P\subset\text{Ob}(\mathcal{A})$ be the set of objects such that $R^iF(X) = 0$ for all $X\in P$ and $i\ne n$, and suppose every object of $\mathcal{A}$ is a quotient of an object in $P$.

Let $G : \mathcal{A}\rightarrow\mathcal{B}$ be the functor $R^nF$. Since $F$ has cohomological dimension $\le n$, $G$ is right exact, and sends exact sequences of objects in $P$ to exact sequences.

Now let $X^\bullet$ be a complex of objects in $P$. Let $I^{\bullet,\bullet}$ be a Cartan-Eilenberg resolution.

Let's fix our attention on a single "column" of the resolution, which we'll write: $$0\rightarrow X\rightarrow I^0\stackrel{d^0}{\rightarrow} I^1\stackrel{d^1}{\rightarrow} I^2\stackrel{d^2}{\rightarrow} \cdots$$ where $X\in P$, and $I^i$ are injective objects of $\mathcal{A}$, and is an injective resolution of $X$. Consider the truncated exact sequence:

$$0\rightarrow X\rightarrow I^0\rightarrow I^1\rightarrow\cdots\rightarrow I^{n-1}\rightarrow\text{Ker}(d^n)\rightarrow 0$$

At the top of page 78 (second part of the proof of Proposition I.7.4), Hartshorne claims that because $X\in P$, and $I^i$ are all injective, then $\text{Ker}(d^n)$ is $F$-acyclic.

This must be obvious, but I can't see why this is true. Any attempts to prove it get stuck when I realize that $\text{Ker}(d^n)$ is just some random subobject of an injective object.


2 Answers 2


This is a standard dimension shifting argument (décalage): split the long exact sequence $$0 \to X \to I^0 \to I^1 \to \ldots \to I^{n-1} \to \ker(d^n) \to 0$$ into short exact sequences $$0 \to J^i \to I^i \to J^{i+1} \to 0.$$ Here, $J^0 = X$ and $J^n = \ker(d^n)$. For each $k$, the associated long exact sequence gives $$\ldots \to R^kF(I^i) \to R^kF(J^{i+1}) \to R^{k+1}F(J^i) \to R^{k+1}F(I^i) \to \ldots.$$ For $k>0$, the outer terms vanish, giving an isomorphism $R^kF(J^{i+1}) \stackrel\sim\to R^{k+1}F(J^i)$. By induction, $$R^kF(J^n) = R^{k+1}F(J^{n-1}) = \ldots = R^{k+n}F(J^0).$$ This vanishes when $k+n \neq n$, i.e. when $k > 0$. $\square$

Remark. The only thing we use is that $R^iF(I) = 0$ for $i > 0$ and $I$ injective. For left exact functors, this is well-known. You should convince yourself it's still true without that assumption (the way Hartshorne sets it up, this is a little hidden, but it's still true more or less by definition). Similarly, the long exact sequence is never stated explicitly, but it's definitely still fine.


For my own benefit, I will basically transcribe Hartshorne's proof here, with some elaborations:

We are trying to prove, that in the situation of the OP, that $RF$ and $LG$ exist, and there is a functorial isomorphism $\psi : RF\cong LG[-n]$.

RF exists - This follows from Lemma I.4.6 and Corollary I.5.3 under the assumptions that $\mathcal{A}$ has enough injectives, and $F$ has finite cohomological dimension.

LG exists - It's easy to see that $P$ satisfies (i)$^*$ and (ii)$^*$ of Lemma I.4.6. To show that it satisfies (iii)$^*$ of Lemma I.4.6, we will show that if $$0\rightarrow X^0\rightarrow X^1\rightarrow X^2\rightarrow\cdots\rightarrow X^n$$ is exact (here $n$ is the cohomological dimension of $F$), with $X^1,X^2,\ldots,X^n\in P$, then $X^0\in P$ as well. For this, we play the same game as in de Bruyn's answer. Let $K^i\subset X^i$ be the kernel of $X^i\rightarrow X^{i+1}$. Then for every $i\ge 1$ we have an exact sequence $$0\rightarrow K^i\rightarrow X^i\rightarrow K^{i+1}\rightarrow 0$$ (where $X^i\in P$) whence, a long exact sequence $$0\rightarrow R^0F(K^i)\rightarrow \underbrace{R^0F(X^i)}_{0}\rightarrow R^0F(K^{i+1})\rightarrow R^1F(K^i)\rightarrow \underbrace{R^1F(X^i)}_{0}\rightarrow\cdots$$ where for $i\ge 1$, $R^jF(X^i) = 0$ for all $j < n$, which implies that $R^jF(K^i)\cong R^{j-1}F(k^{i+1})$ for all $j < n,i\ge 1$. Thus, we have: $R^0F(K^1) = 0$, $R^1F(K^1) = R^0F(K^2) = 0$, $R^2F(K^1)=R^1F(K^2)=R^0F(K^3) = 0$, and so on, showing that $R^jF(K^1) = 0$ for all $j < n$. Of course since $F$ has cohomological dimension $n$, $R^jF(K^1) = 0$ for all $j\ne n$. Since $K^1 = X^0$, this shows that $P$ satisfies (iii)$^*$.

It is also clear from the definition that $P$ satisfies (iv)$^*$ of Cor I.5.3 (w.r.t. $G = R^nF$). Thus, if $L\subset K(\mathcal{A})$ is the triangulated subcategory of complexes in $P$, then $P$ satisfies (the dual of) the conditions of Theorem I.5.1, so $LG$ exists.

To define the morphism $\psi : RF\cong LG[-n]$, we note that $L_{Qis}\rightarrow D(\mathcal{A})$ is an equivalence of categories, so it suffices to define $\psi$ on $L_{Qis}$, or, in fact, by Proposition I.3.4, it suffices to define $\psi$ on $L$. Thus, for every complex $X^\bullet$ of objects of $P$, we must give a morphism in $D(\mathcal{B})$ $$\psi(X^\bullet) : RF(X^\bullet)\rightarrow LG(X^\bullet)[-n]$$ such that for any morphism of complexes $f : X^\bullet\rightarrow Y^\bullet$, there is a commutative diagram $$LG(f)\circ\psi(X^\bullet) = \psi(Y^\bullet)\circ RF(f)$$ To define this morphism, since $\mathcal{A}$ has enough injectives, by Lemma I.7.4, $X^\bullet$ admits a Cartan-Eilenberg resolution - that is, a double complex $C^{\bullet,\bullet}$ of injective objects with a map $X^\bullet\rightarrow C^{\bullet,0}$, such that, if we view $X^\bullet$ as "horizontal", then for every $p\in\mathbb{Z}$, the $p$th column $X^p\rightarrow C^{p,\bullet}$ is a resolution of $X^p$, and the same is true of all kernels, images, and cohomlogy of the horizontal maps.

Consider the truncated complex $C'^{\bullet,\bullet}$ defined by: $$C'^{p,q} = \left\{\begin{array}{ll} C^{p,q} & q < n \\ \text{Ker}(d_2^{p,n}) & q = n \\ 0 & q > n \end{array}\right.$$ Then, for every $p\in\mathbb{Z}$, we have an exact sequence $$0\rightarrow X^p\rightarrow C'^{p,0}\rightarrow C'^{p,1}\rightarrow\cdots\rightarrow C'^{p,n-1}\rightarrow C'^{p,n}\rightarrow 0$$ Where $C'^{p,0},\ldots C'^{p,n-1}$ are all injective, and by de Bruyn's answer to the OP, $C'^{p,n}$ is acyclic, and hence $X^p\rightarrow C'^{p,\bullet}$ gives an acyclic resolution of $X^p$, which can be used to calculate cohomology - that is, we have an isomorphism: $$G(X^p) = R^nF(X^p)\stackrel{\sim}{\longrightarrow} F(C'^{p,n})/\text{Im }F(C'^{p,n-1})$$ From this, we get a map $\alpha^p : F(C'^{p,n})\rightarrow G(X^p)$, whence a map $\alpha : F(C'^{\bullet,\bullet})\rightarrow G(X^\bullet)$ of double complexes, where we view $G(X^\bullet)$ as a double complex concentrated in the $n$th row.

Taking the associated "simple" (or "total") complexes, we get an isomorphism: $$s(\alpha) : F(sC'^{\bullet,\bullet})\stackrel{\sim}{\longrightarrow}G(X^\bullet)[-n]$$ Now, $sC'^{\bullet,\bullet}$ is a complex of $F$-acyclic objects, and by Lemma I.7.5(f), the map of complexes $u : X^\bullet\rightarrow sC'^{\bullet,\bullet}$ is a quasi-isomorphism, and hence we have an isomorphism $$\varphi(u) : RF(X^\bullet)\stackrel{\sim}{\longrightarrow}F(sC'^{\bullet,\bullet})$$

Since $X^\bullet$ is by definition made of $G$-acyclic objects, there is an isomorphism $$\varphi(\text{id}_X) : G(X^\bullet)\stackrel{\sim}{\longrightarrow} LG(X^\bullet)$$ and by composing the maps above, we get a map: $$\psi(X^\bullet) := \varphi(\text{id}_X)[-n]\circ s(\alpha)\circ\varphi(u) \quad:\quad RF(X^\bullet)\rightarrow LG(X^\bullet)[-n]$$

The map $\psi(X^\bullet)$ is an isomorphism for every $X^p$, and since $RF,LG$ are "way out in both directions" (ie, they send complexes of bounded cohomological dimension (in both directions) to complexes with bounded cohomological dimension (in both directions)), by Proposition I.7.1, this implies that $\psi(*)$ defines a natural isomorphism of functors.

Note that $\psi(X^\bullet)$ may depend on the choice of a Cartan-Eilenberg resolution of $X^\bullet$, but nonetheless one may verify that it satisfies the compatibility condition: for any morphism of complexes $f : X^\bullet\rightarrow Y^\bullet$, there is a commutative diagram $$LG(f)\circ\psi(X^\bullet) = \psi(Y^\bullet)\circ RF(f)$$


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