## 13.21 Cartan-Eilenberg resolutions

This section can be expanded. The material can be generalized and applied in more cases. Resolutions need not use injectives and the method also works in the unbounded case in some situations.

Definition 13.21.1. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be a bounded below complex. A Cartan-Eilenberg resolution of $K^\bullet$ is given by a double complex $I^{\bullet , \bullet }$ and a morphism of complexes $\epsilon : K^\bullet \to I^{\bullet , 0}$ with the following properties:

1. There exists a $i \ll 0$ such that $I^{p, q} = 0$ for all $p < i$ and all $q$.

2. We have $I^{p, q} = 0$ if $q < 0$.

3. The complex $I^{p, \bullet }$ is an injective resolution of $K^ p$.

4. The complex $\mathop{\mathrm{Ker}}(d_1^{p, \bullet })$ is an injective resolution of $\mathop{\mathrm{Ker}}(d_ K^ p)$.

5. The complex $\mathop{\mathrm{Im}}(d_1^{p, \bullet })$ is an injective resolution of $\mathop{\mathrm{Im}}(d_ K^ p)$.

6. The complex $H^ p_ I(I^{\bullet , \bullet })$ is an injective resolution of $H^ p(K^\bullet )$.

Lemma 13.21.2. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $K^\bullet$ be a bounded below complex. There exists a Cartan-Eilenberg resolution of $K^\bullet$.

Proof. Suppose that $K^ p = 0$ for $p < n$. Decompose $K^\bullet$ into short exact sequences as follows: Set $Z^ p = \mathop{\mathrm{Ker}}(d^ p)$, $B^ p = \mathop{\mathrm{Im}}(d^{p - 1})$, $H^ p = Z^ p/B^ p$, and consider

$\begin{matrix} 0 \to Z^ n \to K^ n \to B^{n + 1} \to 0 \\ 0 \to B^{n + 1} \to Z^{n + 1} \to H^{n + 1} \to 0 \\ 0 \to Z^{n + 1} \to K^{n + 1} \to B^{n + 2} \to 0 \\ 0 \to B^{n + 2} \to Z^{n + 2} \to H^{n + 2} \to 0 \\ \ldots \end{matrix}$

Set $I^{p, q} = 0$ for $p < n$. Inductively we choose injective resolutions as follows:

1. Choose an injective resolution $Z^ n \to J_ Z^{n, \bullet }$.

2. Using Lemma 13.18.9 choose injective resolutions $K^ n \to I^{n, \bullet }$, $B^{n + 1} \to J_ B^{n + 1, \bullet }$, and an exact sequence of complexes $0 \to J_ Z^{n, \bullet } \to I^{n, \bullet } \to J_ B^{n + 1, \bullet } \to 0$ compatible with the short exact sequence $0 \to Z^ n \to K^ n \to B^{n + 1} \to 0$.

3. Using Lemma 13.18.9 choose injective resolutions $Z^{n + 1} \to J_ Z^{n + 1, \bullet }$, $H^{n + 1} \to J_ H^{n + 1, \bullet }$, and an exact sequence of complexes $0 \to J_ B^{n + 1, \bullet } \to J_ Z^{n + 1, \bullet } \to J_ H^{n + 1, \bullet } \to 0$ compatible with the short exact sequence $0 \to B^{n + 1} \to Z^{n + 1} \to H^{n + 1} \to 0$.

4. Etc.

Taking as maps $d_1^\bullet : I^{p, \bullet } \to I^{p + 1, \bullet }$ the compositions $I^{p, \bullet } \to J_ B^{p + 1, \bullet } \to J_ Z^{p + 1, \bullet } \to I^{p + 1, \bullet }$ everything is clear. $\square$

Lemma 13.21.3. Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor of abelian categories. Let $K^\bullet$ be a bounded below complex of $\mathcal{A}$. Let $I^{\bullet , \bullet }$ be a Cartan-Eilenberg resolution for $K^\bullet$. The spectral sequences $({}'E_ r, {}'d_ r)_{r \geq 0}$ and $({}''E_ r, {}''d_ r)_{r \geq 0}$ associated to the double complex $F(I^{\bullet , \bullet })$ satisfy the relations

${}'E_1^{p, q} = R^ qF(K^ p) \quad \text{and} \quad {}''E_2^{p, q} = R^ pF(H^ q(K^\bullet ))$

Moreover, these spectral sequences are bounded, converge to $H^*(RF(K^\bullet ))$, and the associated induced filtrations on $H^ n(RF(K^\bullet ))$ are finite.

Proof. We will use the following remarks without further mention:

1. As $I^{p, \bullet }$ is an injective resolution of $K^ p$ we see that $RF$ is defined at $K^ p[0]$ with value $F(I^{p, \bullet })$.

2. As $H^ p_ I(I^{\bullet , \bullet })$ is an injective resolution of $H^ p(K^\bullet )$ the derived functor $RF$ is defined at $H^ p(K^\bullet )[0]$ with value $F(H^ p_ I(I^{\bullet , \bullet }))$.

3. By Homology, Lemma 12.25.4 the total complex $\text{Tot}(I^{\bullet , \bullet })$ is an injective resolution of $K^\bullet$. Hence $RF$ is defined at $K^\bullet$ with value $F(\text{Tot}(I^{\bullet , \bullet }))$.

Consider the two spectral sequences associated to the double complex $L^{\bullet , \bullet } = F(I^{\bullet , \bullet })$, see Homology, Lemma 12.25.1. These are both bounded, converge to $H^*(\text{Tot}(L^{\bullet , \bullet }))$, and induce finite filtrations on $H^ n(\text{Tot}(L^{\bullet , \bullet }))$, see Homology, Lemma 12.25.3. Since $\text{Tot}(L^{\bullet , \bullet }) = \text{Tot}(F(I^{\bullet , \bullet })) = F(\text{Tot}(I^{\bullet , \bullet }))$ computes $H^ n(RF(K^\bullet ))$ we find the final assertion of the lemma holds true.

Computation of the first spectral sequence. We have ${}'E_1^{p, q} = H^ q(L^{p, \bullet })$ in other words

${}'E_1^{p, q} = H^ q(F(I^{p, \bullet })) = R^ qF(K^ p)$

as desired. Observe for later use that the maps ${}'d_1^{p, q} : {}'E_1^{p, q} \to {}'E_1^{p + 1, q}$ are the maps $R^ qF(K^ p) \to R^ qF(K^{p + 1})$ induced by $K^ p \to K^{p + 1}$ and the fact that $R^ qF$ is a functor.

Computation of the second spectral sequence. We have ${}''E_1^{p, q} = H^ q(L^{\bullet , p}) = H^ q(F(I^{\bullet , p}))$. Note that the complex $I^{\bullet , p}$ is bounded below, consists of injectives, and moreover each kernel, image, and cohomology group of the differentials is an injective object of $\mathcal{A}$. Hence we can split the differentials, i.e., each differential is a split surjection onto a direct summand. It follows that the same is true after applying $F$. Hence ${}''E_1^{p, q} = F(H^ q(I^{\bullet , p})) = F(H^ q_ I(I^{\bullet , p}))$. The differentials on this are $(-1)^ q$ times $F$ applied to the differential of the complex $H^ p_ I(I^{\bullet , \bullet })$ which is an injective resolution of $H^ p(K^\bullet )$. Hence the description of the $E_2$ terms. $\square$

Remark 13.21.4. The spectral sequences of Lemma 13.21.3 are functorial in the complex $K^\bullet$. This follows from functoriality properties of Cartan-Eilenberg resolutions. On the other hand, they are both examples of a more general spectral sequence which may be associated to a filtered complex of $\mathcal{A}$. The functoriality will follow from its construction. We will return to this in the section on the filtered derived category, see Remark 13.26.15.

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