The Stacks project

Proof. Let $(A_ n)$ be an arbitrary inverse system. Let $(B_ n)$ be the inverse system with

and transition maps given by projections. Let $A_ n \to B_ n$ be given by $(1, f_ n, f_{n - 1} \circ f_ n, \ldots , f_2 \circ \ldots \circ f_ n)$ where $f_ i : A_ i \to A_{i - 1}$ are the transition maps. In this way we see that every inverse system is a subobject of a ML system (Homology, Section 12.31). It follows from Derived Categories, Lemma 13.15.6 using Homology, Lemma 12.31.3 that every ML system is right acyclic for $\mathop{\mathrm{lim}}\nolimits $, i.e., (3) holds. This already implies that $RF$ is defined on $D^+(\textit{Ab}(\mathbf{N}))$, see Derived Categories, Proposition 13.16.8. Set $C_ n = A_{n - 1} \oplus \ldots \oplus A_1$ for $n > 1$ and $C_1 = 0$ with transition maps given by projections as well. Then there is a short exact sequence of inverse systems $0 \to (A_ n) \to (B_ n) \to (C_ n) \to 0$ where $B_ n \to C_ n$ is given by $(x_ i) \mapsto (x_ i - f_{i + 1}(x_{i + 1}))$. Since $(C_ n)$ is ML as well, we conclude that (2) holds (by proposition reference above) which also implies (1). Finally, this implies by Derived Categories, Lemma 13.32.2 that $R\mathop{\mathrm{lim}}\nolimits $ is in fact defined on all of $D(\textit{Ab}(\mathbf{N}))$. In fact, the proof of Derived Categories, Lemma 13.32.2 proceeds by proving assertions (4) and (5). $\square$

Proof. Follows from the vanishing in Lemma 15.86.1. $\square$

Proof. Let $K \in D(\textit{Ab}(\mathbf{N}))$ be the object represented by the system of complexes whose $n$th constituent is the complex $A^{-2}_ n \to A^{-1}_ n \to A^0_ n \to A^1_ n$. We will compute $H^0(R\mathop{\mathrm{lim}}\nolimits K)$ using both spectral sequences¹ of Derived Categories, Lemma 13.21.3. The first has $E_1$-page

with horizontal differentials and all higher differentials are zero. The second has $E_2$ page

and degenerates at this point. The result follows. $\square$

Proof. This follows from Derived Categories, Definition 13.34.1 and Lemma 13.4.2, and the description of $\mathop{\mathrm{lim}}\nolimits $ and $R^1\mathop{\mathrm{lim}}\nolimits $ in Lemma 15.86.1 above. $\square$

Proof. This follows from Derived Categories, Definition 13.33.1 and Lemma 13.4.2, and the description of $\mathop{\mathrm{lim}}\nolimits $ and $R^1\mathop{\mathrm{lim}}\nolimits $ in Lemma 15.86.1 above. $\square$

Proof. Suppose that for each $p$ the inverse system $(K_ n^ p)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits $. By Lemma 15.86.1 this gives a short exact sequence

for each $p$. Since the complex consisting of $\mathop{\mathrm{lim}}\nolimits _ n K^ p_ n$ computes $R\mathop{\mathrm{lim}}\nolimits K$ by Lemma 15.86.1 we see that the lemma holds in this case.

Next, assume $K = (K_ n^\bullet )$ is general. By Lemma 15.86.1 there is a quasi-isomorphism $K \to L$ in $D(\textit{Ab}(\mathbf{N}))$ such that $(L_ n^ p)$ is acyclic for each $p$. Then $\prod K_ n^\bullet $ is quasi-isomorphic to $\prod L_ n^\bullet $ as products are exact in $\textit{Ab}$, whence the result for $L$ (proved above) implies the result for $K$. $\square$

Proof. The long exact sequence of the distinguished triangle is

The map in the middle has kernel $\mathop{\mathrm{lim}}\nolimits _ n H^ p(K_ n^\bullet )$ by its explicit description given in the lemma. The cokernel of this map is $R^1\mathop{\mathrm{lim}}\nolimits _ n H^ p(K_ n^\bullet )$ by Lemma 15.86.1. $\square$

Proof. Namely, let $M_1^\bullet $ be a complex of abelian groups representing $K_1$. Suppose we have constructed $M_ e^\bullet \to M_{e - 1}^\bullet \to \ldots \to M_1^\bullet $ and maps $\psi _ i : M_ i^\bullet \to K_ i$ such that the diagrams in the statement of the lemma commute for all $n < e$. Then we consider the diagram

in $D(\textit{Ab})$. By the definition of morphisms in $D(\textit{Ab})$ we can find a complex $M_{n + 1}^\bullet $ of abelian groups, an isomorphism $M_{n + 1}^\bullet \to K_{n + 1}$ in $D(\textit{Ab})$, and a morphism of complexes $M_{n + 1}^\bullet \to M_ n^\bullet $ representing the composition

in $D(\textit{Ab})$. Thus the lemma holds by induction. $\square$

Proof. The assumption in particular implies that the pro-objects $H^ p(E_ n)$ and $H^ p(D_ n)$ are isomorphic. By the short exact sequences of Lemma 15.86.8 it suffices to show that given a map $(A_ n) \to (B_ n)$ of inverse systems of abelian groupsc which induces an isomorphism of pro-objects, then $\mathop{\mathrm{lim}}\nolimits A_ n \cong \mathop{\mathrm{lim}}\nolimits B_ n$ and $R^1\mathop{\mathrm{lim}}\nolimits A_ n \cong R^1\mathop{\mathrm{lim}}\nolimits B_ n$.

The assumption implies there are $1 \leq m_1 < m_2 < m_3 < \ldots $ and maps $\varphi _ n : B_{m_ n} \to A_ n$ such that $(\varphi _ n) : (B_{m_ n}) \to (A_ n)$ is a map of systems which is inverse to the given map $\psi = (\psi _ n) : (A_ n) \to (B_ n)$ as a morphism of pro-objects. What this means is that (after possibly replacing $m_ n$ by larger integers) we may assume that the compositions $A_{m_ n} \to B_{m_ n} \to A_ n$ and $B_{m_ n} \to A_ n \to B_ n$ are equal to the transition maps of the inverse systems. Now, if $(b_ n) \in \mathop{\mathrm{lim}}\nolimits B_ n$ we can set $a_ n = \varphi _{m_ n}(b_{m_ n})$. This defines an inverse $\mathop{\mathrm{lim}}\nolimits B_ n \to \mathop{\mathrm{lim}}\nolimits A_ n$ (computation omitted). Let us use the cokernel of the map

as an avatar of $R^1\mathop{\mathrm{lim}}\nolimits B_ n$ (Lemma 15.86.1). Any element in this cokernel can be represented by an element $(b_ i)$ with $b_ i = 0$ if $i \not= m_ n$ for some $n$ (computation omitted). We can define a map $R^1\mathop{\mathrm{lim}}\nolimits B_ n \to R^1\mathop{\mathrm{lim}}\nolimits A_ n$ by mapping the class of such a special element $(b_ n)$ to the class of $(\varphi _ n(b_{m_ n}))$. We omit the verification this map is inverse to the map $R^1\mathop{\mathrm{lim}}\nolimits A_ n \to R^1\mathop{\mathrm{lim}}\nolimits B_ n$. $\square$

Proof. Set $B = \bigoplus _{i \in \mathbf{N}} (A_ n)$ and hence $B = (B_ n)$ with $B_ n = \bigoplus _{i \in \mathbf{N}} A_ n$. If $(A_ n)$ is ML, then $B$ is ML and hence $R^1\mathop{\mathrm{lim}}\nolimits A_ n = 0$ and $R^1\mathop{\mathrm{lim}}\nolimits B_ n = 0$ by Lemma 15.86.1.

Conversely, assume $(A_ n)$ is not ML. Then we can pick an $m$ and a sequence of integers $m < m_1 < m_2 < \ldots $ and elements $x_ i \in A_{m_ i}$ whose image $y_ i \in A_ m$ is not in the image of $A_{m_ i + 1} \to A_ m$. We will use the elements $x_ i$ and $y_ i$ to show that $R^1\mathop{\mathrm{lim}}\nolimits B_ n \not= 0$ in two ways. This will finish the proof of the lemma.

First proof. Set $C = (C_ n)$ with $C_ n = \prod _{i \in \mathbf{N}} A_ n$. There is a canonical injective map $B_ n \to C_ n$ with cokernel $Q_ n$. Set $Q = (Q_ n)$. We may and do think of elements $q_ n$ of $Q_ n$ as sequences of elements $q_ n = (q_{n, 1}, q_{n, 2}, \ldots )$ with $q_{n, i} \in A_ n$ modulo sequences whose tail is zero (in other words, we identify sequences which differ in finitely many places). We have a short exact sequence of inverse systems

Consider the element $q_ n \in Q_ n$ given by

Then it is clear that $q_{n + 1}$ maps to $q_ n$. Hence we obtain $q = (q_ n) \in \mathop{\mathrm{lim}}\nolimits Q_ n$. On the other hand, we claim that $q$ is not in the image of $\mathop{\mathrm{lim}}\nolimits C_ n \to \mathop{\mathrm{lim}}\nolimits Q_ n$. Namely, say that $c = (c_ n)$ maps to $q$. Then we can write $c_ n = (c_{n, i})$ and since $c_{n', i} \mapsto c_{n, i}$ for $n' \geq n$, we see that $c_{n, i} \in \mathop{\mathrm{Im}}(C_{n'} \to C_ n)$ for all $n, i, n' \geq n$. In particular, the image of $c_{m, i}$ in $A_ m$ is in $\mathop{\mathrm{Im}}(A_{m_ i + 1} \to A_ m)$ whence cannot be equal to $y_ i$. Thus $c_ m$ and $q_ m = (y_1, y_2, y_3, \ldots )$ differ in infinitely many spots, which is a contradiction. Considering the long exact cohomology sequence

we conclude that the last group is nonzero as desired.

Second proof. For $n' \geq n$ we denote $A_{n, n'} = \mathop{\mathrm{Im}}(A_{n'} \to A_ n)$. Then we have $y_ i \in A_ m$, $y_ i \not\in A_{m, m_ i + 1}$. Let $\xi = (\xi _ n) \in \prod B_ n$ be the element with $\xi _ n = 0$ unless $n = m_ i$ and $\xi _{m_ i} = (0, \ldots , 0, x_ i, 0, \ldots )$ with $x_ i$ placed in the $i$th summand. We claim that $\xi $ is not in the image of the map $\prod B_ n \to \prod B_ n$ of Lemma 15.86.1. This shows that $R^1\mathop{\mathrm{lim}}\nolimits B_ n$ is nonzero and finishes the proof. Namely, suppose that $\xi $ is the image of $\eta = (z_1, z_2, \ldots )$ with $z_ n = \sum z_{n, i} \in \bigoplus _ i A_ n$. Observe that $x_ i = z_{m_ i, i} \bmod A_{m_ i, m_ i + 1}$. Then $z_{m_ i - 1, i}$ is the image of $z_{m_ i, i}$ under $A_{m_ i} \to A_{m_ i - 1}$, and so on, and we conclude that $z_{m, i}$ is the image of $z_{m_ i, i}$ under $A_{m_ i} \to A_ m$. We conclude that $z_{m, i}$ is congruent to $y_ i$ modulo $A_{m, m_ i + 1}$. In particular $z_{m, i} \not= 0$. This is impossible as $\sum z_{m, i} \in \bigoplus _ i A_ m$ hence only a finite number of $z_{m, i}$ can be nonzero. $\square$

Proof. This follows from Lemma 15.86.12, the fact that taking infinite direct sums is exact, and the long exact sequence of cohomology associated to $R\mathop{\mathrm{lim}}\nolimits $. $\square$

Proof. It follows from Lemma 15.86.11 that (1) implies (2). Assume (2). Then $(A_ n)$ is ML by Lemma 15.86.12. For $m \geq n$ let $A_{n, m} = \mathop{\mathrm{Im}}(A_ m \to A_ n)$ so that $A_ n = A_{n, n} \supset A_{n, n + 1} \supset \ldots $. Note that $(A_ n)$ is zero as a pro-object if and only if for every $n$ there is an $m \geq n$ such that $A_{n, m} = 0$. Note that $(A_ n)$ is ML if and only if for every $n$ there is an $m_ n \geq n$ such that $A_{n, m} = A_{n, m + 1} = \ldots $. In the ML case it is clear that $\mathop{\mathrm{lim}}\nolimits A_ n = 0$ implies that $A_{n, m_ n} = 0$ because the maps $A_{n + 1, m_{n + 1}} \to A_{n, m}$ are surjective. This finishes the proof. $\square$