Remark on equality of cohomology groups (Milne, Remark 1.14)

Question

I'm having troubles with Remark 1.14, page 69 of these notes. Explicitly, if $$ 0 \to M \to J^0 \to J^1 \to \cdots $$ is an exact sequence of $ G $-modules, such that $ H^s(G, J^r) = 0$ for all $ s > 0 $ and all $ r $, then $ H^r(G, M) = H^r(J^{\bullet G}) $. I'm not sure if I'm missing something obvious or if it's not that easy to show, since the author did not give any explanation. Thank you for your help.

Answer 1

This is a standard result in homological algebra as stated in the comments. The statement is as follows.

Let $F: \mathscr A\to \mathscr B$ be a left exact functor between abelian categories with enough injectives. An object $X$ is $F$-acyclic if $R^iF(X)=0$ for $i>0$. If $X\to Y^*$ is a resolution of $X$ by $F$-acyclic objects, then $H^i(FY^*) = R^iF(X)$ for each $i$.

To prove this, one produces a double complex of injectives, which I will denote by $J^{i,j}$ ($i\geqslant -1, j\geqslant -1$), with (at least) the following properties.

1. For each fixed $i$, $J^i = J^{i,-1} \to J^{i,*}$ is an injective resolution ($i=-1$ gives the starting column for $X$), in particular $X$ is at position $(-1,-1)$.

2. For each $j$, the complex $J^{-1,j} \to J^{*,j}$ is an injective resolution.

The construction uses the Horseshoe lemma in a smart way, but otherwise is very elementary. The point is to consider it a sort of resolution (what is called a Cartan--Eilenberg resolution) of the complex $C^{*,-1}: M\to J^*$:

$$C^{*,-1} = J^{*,-1} \to J^{*,0}\to J^{*,1}\to J^{*,2}\to \cdots$$

For a more precise definition, you may consult Theorem 10.45 and the preceding discussion in Rotman's Homological Algebra (2nd edition). Ultimately one obtains the Grothendieck spectral sequence pursuing this idea and using the spectral sequence of a double complex.

In particular one has two maps of double complexes: one $\varepsilon$ from the injective resolution $J^{-1,*}$ to the remaining part $J_+$ of the complex (this diagram does not involve the coordinates with $i,j=-1$) given by the horizontal differential, and a map $\eta$ from $J^{*,-1}$ to $J_+$ given by the vertical one. Applying $F$, one gets the same diagrams but with $F$ about.

Now, since the $J^{i}$ are $F$-acyclic, this $F$-complex plus the row of $FJ^{*,-1}$ has acyclic columns, so its totalization is acyclic. Similarly, since injective objects are $F$-acyclic, this $F$-complex plus the column of $FJ^{-1,*}$ has acyclic rows, so its totalization is acyclic. Here we use the "Acyclic Assembly Lemma" of Weibel, which is a very simple version of the spectral sequence of a double complex.

Viewing the horizontal (resp. vertical) maps as morphisms $\varepsilon$ (resp $\varepsilon'$) of doble complexes, one notices that the cones of the maps $\operatorname{Tot}(F\varepsilon)$ and $\operatorname{Tot}(F\varepsilon')$ are both isomorphic to the totalization of the two augmented $F$-complexes, and since they are acyclic, both these maps are quasi-isomorphisms.

Now the domain of $\operatorname{Tot}(F\varepsilon)$ is just the complex that computes $R^iFM$, while the domain of $\operatorname{Tot}(F\varepsilon')$ is the complex that computes $H^i(FI^*)$, and both these complexes are quasi-isomorphic to $\operatorname{Tot}(FJ_+)$ through the maps above. This constructs the desired isomorphisms:

$$R^*FM \leftarrow H^*(\operatorname{Tot}(FJ_+))\rightarrow H^*(FI^*) $$