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 15.87.1. $\square$

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. Please see Categories, Example 4.22.6 for a discussion of morphisms of pro-systems. The map $\varphi $ is given by $1 \leq m_1 < m_2 < m_3 < \ldots $ and maps $\varphi _ n : A_{m_ n} \to B_ n$ compatible with transition maps. Set $\varphi ', \varphi '' : \prod A_ n \to \prod B_ n$ equal to $\varphi '((a_ n)) = (\varphi _ n(a_{m_ n}))$ and

where each occurence of $f$ denotes a suitable transition map of the inverse system $(A_ n)$. Then the diagram

where the horizontal arrows are as in Lemma 15.87.1 is commutative. In this way we obtain the desired maps. The construction is functorial in the sense that if we're given an inverse system $(C_ n)$ and $1 \leq m'_1 < m'_2 < m'_3 < \ldots $ and maps $\psi _ n : B_{m'_ n} \to C_ n$ compatible with transitition maps, then $(\psi \circ \varphi )' = \psi ' \circ \varphi '$ and $(\psi \circ \varphi )'' = \psi '' \circ \varphi ''$ where the composition $\psi \circ \varphi $ refers to the integers $1 \leq m_{m'_1} < m_{m'_2} < \ldots $ and the maps $\psi _ n \circ \varphi _{m'_ n} : A_{m_{m'_ n}} \to C_ n$. We will show that if $B_ n = A_ n$ and $\varphi _ n$ is the transition map, then the resulting maps are the identity maps. This will both prove that the construction is independent of the choice of the representative $(m_ n, \varphi _ n)$ of $\varphi $ and the final statement of the lemma.

Thus we let $B_ n = A_ n$ and $\varphi _ n$ be the transition map. Let $(a_ n) \in \mathop{\mathrm{lim}}\nolimits A_ n$ be an element. Then $\varphi '((a_ n))$ is the element which has in degree $n$ the image of $a_{m_ n}$ which is equal to $a_ n$. This proves the statement for $\mathop{\mathrm{lim}}\nolimits A_ n$. Let $\xi \in R^1\mathop{\mathrm{lim}}\nolimits A_ n$ be the class of the element $(a_ n)$ in $\prod A_ n$. Consider the element $(b_ n)$ with

if $m_{n - 1} < i < m_ n$ and $0$ if $i = m_ n$ for some $n$. Then $(a'_ n) = (a_ n) - \delta ((b_ n))$ defines the same class in $R^1\mathop{\mathrm{lim}}\nolimits A_ n$ and a computation shows that $a'_ i = 0$ unless $i = m_ n$ for some $n$. Thus we may and do assume $a_ i = 0$ unless $i = m_ n$ for some $n$. Note that

with nonzero entries in spots $j$ and $m_ j$, is the image under $\delta $ of

with nonzero entries in spots $j + 1, \ldots , m_ j$. The sum $c = \sum c_ j$ makes sense in $\prod A_ n$. Recalling that $\varphi ''((a_ n)) = (a_{m_1}, a_{m_2}, \ldots )$ we see that

and the proof is complete. $\square$

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 15.87.1 above. $\square$

Proof. We obtain an arrow $K \to M$ fitting into a morphism

of defining distinguished triangles by Derived Categories, Remark 13.34.4. Thus, for every object $L$ of $\mathcal{D}$ we obtain a map of short exact sequences

see Lemma 15.87.5 and its proof. By Lemma 15.87.4 the left and right vertical arrows are isomorphisms². Thus the middle arrow is an isomorphism. By the Yoneda lemma we see that the map $K \to M$ is an isomorphism. $\square$

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 15.87.1 above. $\square$

Proof. Suppose that for each $p$ the inverse system $(K_ n^ p)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits $. By Lemma 15.87.1 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 15.87.1 we see that the lemma holds in this case.

Next, assume $K = (K_ n^\bullet )$ is general. By Lemma 15.87.1 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 15.87.1. $\square$

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. The result follows from the short exact sequences of Lemma 15.87.10 and Lemma 15.87.4. $\square$

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 15.87.1.

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 15.87.1. 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 15.87.14, 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 15.87.4 that (1) implies (2). Assume (2). Then $(A_ n)$ is ML by Lemma 15.87.14. 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. $\square$