
## 15.77 Rlim of abelian groups

We briefly discuss $R\mathop{\mathrm{lim}}\nolimits$ on abelian groups. In this section we will denote $\textit{Ab}(\mathbf{N})$ the abelian category of inverse systems of abelian groups. The notation is compatible with the notation for sheaves of abelian groups on a site, as an inverse system of abelian groups is the same thing as a sheaf of groups on the category $\mathbf{N}$ (with a unique morphism $i \to j$ if $i \leq j$), see Remark 15.77.5. Many of the arguments in this section duplicate the arguments used to construct the cohomological machinery for sheaves of abelian groups on sites.

Lemma 15.77.1. The functor $\mathop{\mathrm{lim}}\nolimits : \textit{Ab}(\mathbf{N}) \to \textit{Ab}$ has a right derived functor

15.77.1.1
$$\label{more-algebra-equation-Rlim} R\mathop{\mathrm{lim}}\nolimits : D(\textit{Ab}(\mathbf{N})) \longrightarrow D(\textit{Ab})$$

As usual we set $R^ p\mathop{\mathrm{lim}}\nolimits (K) = H^ p(R\mathop{\mathrm{lim}}\nolimits (K))$. Moreover, we have

1. for any $(A_ n)$ in $\textit{Ab}(\mathbf{N})$ we have $R^ p\mathop{\mathrm{lim}}\nolimits A_ n = 0$ for $p > 1$,

2. the object $R\mathop{\mathrm{lim}}\nolimits A_ n$ of $D(\textit{Ab})$ is represented by the complex

$\prod A_ n \to \prod A_ n,\quad (x_ n) \mapsto (x_ n - f_{n + 1}(x_{n + 1}))$

sitting in degrees $0$ and $1$,

3. if $(A_ n)$ is ML, then $R^1\mathop{\mathrm{lim}}\nolimits A_ n = 0$, i.e., $(A_ n)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits$,

4. every $K^\bullet \in D(\textit{Ab}(\mathbf{N}))$ is quasi-isomorphic to a complex whose terms are right acyclic for $\mathop{\mathrm{lim}}\nolimits$, and

5. if each $K^ p = (K^ p_ n)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits$, i.e., of $R^1\mathop{\mathrm{lim}}\nolimits _ n K^ p_ n = 0$, then $R\mathop{\mathrm{lim}}\nolimits K$ is represented by the complex whose term in degree $p$ is $\mathop{\mathrm{lim}}\nolimits _ n K_ n^ p$.

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

$B_ n = A_ n \oplus A_{n - 1} \oplus \ldots \oplus A_1$

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 12.28). It follows from Derived Categories, Lemma 13.16.6 using Homology, Lemma 12.28.3 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 13.17.8. 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 13.30.2 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 13.30.2 proceeds by proving assertions (4) and (5). $\square$

We give two simple applications. The first is the “correct” formulation of Homology, Lemma 12.28.7.

$(A^{-2}_ n \to A^{-1}_ n \to A^0_ n \to A^1_ n)$

be an inverse system of complexes of abelian groups and denote $A^{-2} \to A^{-1} \to A^0 \to A^1$ its limit. Denote $(H_ n^{-1})$, $(H_ n^0)$ the inverse systems of cohomologies, and denote $H^{-1}$, $H^0$ the cohomologies of $A^{-2} \to A^{-1} \to A^0 \to A^1$. If

1. $(A^{-2}_ n)$ and $(A^{-1}_ n)$ have vanishing $R^1\mathop{\mathrm{lim}}\nolimits$,

2. $(H^{-1}_ n)$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits$,

then $H^0 = \mathop{\mathrm{lim}}\nolimits H_ n^0$.

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

$\begin{matrix} 0 & 0 & R^1\mathop{\mathrm{lim}}\nolimits A^0_ n & R^1\mathop{\mathrm{lim}}\nolimits A^1_ n \\ A^{-2} & A^{-1} & A^0 & A^1 \end{matrix}$

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

$\begin{matrix} R^1\mathop{\mathrm{lim}}\nolimits H^{-2}_ n & 0 & R^1\mathop{\mathrm{lim}}\nolimits H^0_ n & R^1 \mathop{\mathrm{lim}}\nolimits H^1_ n \\ \mathop{\mathrm{lim}}\nolimits H^{-2}_ n & \mathop{\mathrm{lim}}\nolimits H^{-1}_ n & \mathop{\mathrm{lim}}\nolimits H^0_ n & \mathop{\mathrm{lim}}\nolimits H^1_ n \end{matrix}$

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

Lemma 15.77.3. Let $\mathcal{D}$ be a triangulated category. Let $(K_ n)$ be an inverse system of objects of $\mathcal{D}$. Let $K$ be a derived limit of the system $(K_ n)$. Then for every $L$ in $\mathcal{D}$ we have a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(L, K_ n[-1]) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(L, K) \to \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(L, K_ n) \to 0$

Proof. This follows from Derived Categories, Definition 13.32.1 and Lemma 13.4.2, and the description of $\mathop{\mathrm{lim}}\nolimits$ and $R^1\mathop{\mathrm{lim}}\nolimits$ in Lemma 15.77.1 above. $\square$

Lemma 15.77.4. Let $\mathcal{D}$ be a triangulated category. Let $(K_ n)$ be a system of objects of $\mathcal{D}$. Let $K$ be a derived colimit of the system $(K_ n)$. Then for every $L$ in $\mathcal{D}$ we have a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(K_ n, L[-1]) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(K, L) \to \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(K_ n, L) \to 0$

Proof. This follows from Derived Categories, Definition 13.31.1 and Lemma 13.4.2, and the description of $\mathop{\mathrm{lim}}\nolimits$ and $R^1\mathop{\mathrm{lim}}\nolimits$ in Lemma 15.77.1 above. $\square$

Remark 15.77.5 (Rlim as cohomology). Consider the category $\mathbf{N}$ whose objects are natural numbers and whose morphisms are unique arrows $i \to j$ if $j \geq i$. Endow $\mathbf{N}$ with the chaotic topology (Sites, Example 7.6.6) so that a sheaf $\mathcal{F}$ is the same thing as an inverse system

$\mathcal{F}_1 \leftarrow \mathcal{F}_2 \leftarrow \mathcal{F}_3 \leftarrow \ldots$

of sets over $\mathbf{N}$. Note that $\Gamma (\mathbf{N}, \mathcal{F}) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. For an inverse system of abelian groups $\mathcal{F}_ n$ we have

$R^ p\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n = H^ p(\mathbf{N}, \mathcal{F})$

because both sides are the higher right derived functors of $\mathcal{F} \mapsto \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n = H^0(\mathbf{N}, \mathcal{F})$. Thus the existence of $R\mathop{\mathrm{lim}}\nolimits$ also follows from the general material in Cohomology on Sites, Sections 21.3 and 21.20.

The products in the following lemma can be seen as termwise products of complexes or as products in the derived category $D(\textit{Ab})$, see Derived Categories, Lemma 13.32.2.

Lemma 15.77.6. Let $K = (K_ n^\bullet )$ be an object of $D(\textit{Ab}(\mathbf{N}))$. There exists a canonical distinguished triangle

$R\mathop{\mathrm{lim}}\nolimits K \to \prod \nolimits _ n K_ n^\bullet \to \prod \nolimits _ n K_ n^\bullet \to R\mathop{\mathrm{lim}}\nolimits K[1]$

in $D(\textit{Ab})$. In other words, $R\mathop{\mathrm{lim}}\nolimits K$ is a derived limit of the inverse system $(K_ n^\bullet )$ of $D(\textit{Ab})$, see Derived Categories, Definition 13.32.1.

Proof. Suppose that for each $p$ the inverse system $(K_ n^ p)$ is right acyclic for $\mathop{\mathrm{lim}}\nolimits$. By Lemma 15.77.1 this gives a short exact sequence

$0 \to \mathop{\mathrm{lim}}\nolimits _ n K^ p_ n \to \prod \nolimits _ n K^ p_ n \to \prod \nolimits _ n K^ p_ n \to 0$

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 15.77.1 we see that the lemma holds in this case.

Next, assume $K = (K_ n^\bullet )$ is general. By Lemma 15.77.1 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 $K$. $\square$

Lemma 15.77.7. With notation as in Lemma 15.77.6 the long exact cohomology sequence associated to the distinguished triangle breaks up into short exact sequences

$0 \to R^1\mathop{\mathrm{lim}}\nolimits _ n H^{p - 1}(K_ n^\bullet ) \to H^ p(R\mathop{\mathrm{lim}}\nolimits K) \to \mathop{\mathrm{lim}}\nolimits _ n H^ p(K_ n^\bullet ) \to 0$

Proof. The long exact sequence of the distinguished triangle is

$\ldots \to H^ p(R\mathop{\mathrm{lim}}\nolimits K) \to \prod \nolimits _ n H^ p(K_ n^\bullet ) \to \prod \nolimits _ n H^ p(K_ n^\bullet ) \to H^{p + 1}(R\mathop{\mathrm{lim}}\nolimits K) \to \ldots$

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 15.77.1. $\square$

Warning. An object of $D(\textit{Ab}(\mathbf{N}))$ is a complex of inverse systems of abelian groups. You can also think of this as an inverse system $(K_ n^\bullet )$ of complexes. However, this is not the same thing as an inverse system of objects of $D(\textit{Ab})$; the following lemma and remark explain the difference.

Lemma 15.77.8. Let $(K_ n)$ be an inverse system of objects of $D(\textit{Ab})$. Then there exists an object $M = (M_ n^\bullet )$ of $D(\textit{Ab}(\mathbf{N}))$ and isomorphisms $M_ n^\bullet \to K_ n$ in $D(\textit{Ab})$ such that the diagrams

$\xymatrix{ M_{n + 1}^\bullet \ar[d] \ar[r] & M_ n^\bullet \ar[d] \\ K_{n + 1} \ar[r] & K_ n }$

commute in $D(\textit{Ab})$.

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

$\xymatrix{ & M_ n^\bullet \ar[d]^{\psi _ n} \\ K_{n + 1} \ar[r] & K_ n }$

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

$K_{n + 1} \to K_ n \xrightarrow {\psi _ n^{-1}} M_ n^\bullet$

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

Remark 15.77.9. Let $(K_ n)$ be an inverse system of objects of $D(\textit{Ab})$. Let $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ be a derived limit of this system (see Derived Categories, Section 13.32). Such a derived limit exists because $D(\textit{Ab})$ has countable products (Derived Categories, Lemma 13.32.2). By Lemma 15.77.8 we can also lift $(K_ n)$ to an object $M$ of $D(\mathbf{N})$. Then $K \cong R\mathop{\mathrm{lim}}\nolimits M$ where $R\mathop{\mathrm{lim}}\nolimits$ is the functor (15.77.1.1) because $R\mathop{\mathrm{lim}}\nolimits M$ is also a derived limit of the system $(K_ n)$ by Lemma 15.77.6. Thus, although there may be many isomorphism classes of lifts $M$ of the system $(K_ n)$, the isomorphism type of $R\mathop{\mathrm{lim}}\nolimits M$ is independent of the choice because it is isomorphic to the derived limit $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ of the system. Thus we may apply results on $R\mathop{\mathrm{lim}}\nolimits$ proved in this section to derived limits. For example, for every $p \in \mathbf{Z}$ there is a canonical short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(K_ n) \to H^ p(K) \to \mathop{\mathrm{lim}}\nolimits H^ p(K_ n) \to 0$

because we may apply Lemma 15.77.6 to $M$. This can also been seen directly, without invoking the existence of $M$, by applying the argument of the proof of Lemma 15.77.6 to the (defining) distinguished triangle $K \to \prod K_ n \to \prod K_ n \to K[1]$.

Lemma 15.77.10. Let $E \to D$ be a morphism of $D(\textit{Ab}(\mathbf{N}))$. Let $(E_ n)$, resp. $(D_ n)$ be the system of objects of $D(\textit{Ab})$ associated to $E$, resp. $D$. If $(E_ n) \to (D_ n)$ is an isomorphism of pro-objects, then $R\mathop{\mathrm{lim}}\nolimits E \to R\mathop{\mathrm{lim}}\nolimits D$ is an isomorphism in $D(\textit{Ab})$.

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 15.77.7 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

$\prod B_ n \longrightarrow \prod B_ n$

as an avatar of $R^1\mathop{\mathrm{lim}}\nolimits B_ n$ (Lemma 15.77.1). 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 $R^1\mathop{\mathrm{lim}}\nolimits A_ n \to R^1\mathop{\mathrm{lim}}\nolimits B_ n$. $\square$

Lemma 15.77.11 (Emmanouil). Let $(A_ n)$ be an inverse system of abelian groups. The following are equivalent

1. $(A_ n)$ is Mittag-Leffler,

2. $R^1\mathop{\mathrm{lim}}\nolimits A_ n = 0$ and the same holds for $\bigoplus _{i \in \mathbf{N}} (A_ n)$.

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 15.77.1.

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

$0 \to (B_ n) \to (C_ n) \to (Q_ n) \to 0$

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

$q_{n, i} = \left\{ \begin{matrix} \text{image of }x_ i & \text{if} & m_ i \geq n \\ 0 & \text{else} \end{matrix} \right.$

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

$0 \to \mathop{\mathrm{lim}}\nolimits B_ n \to \mathop{\mathrm{lim}}\nolimits C_ n \to \mathop{\mathrm{lim}}\nolimits Q_ n \to R^1\mathop{\mathrm{lim}}\nolimits B_ n$

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 15.77.1. 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. $\square$

$0 \to (A_ i) \to (B_ i) \to (C_ i) \to 0$

be a short exact sequence of inverse systems of abelian groups. If $(A_ i)$ and $(C_ i)$ are ML, then so is $(B_ i)$.

Proof. This follows from Lemma 15.77.11, the fact that taking infinite direct sums is exact, and the long exact sequence of cohomology associated to $R\mathop{\mathrm{lim}}\nolimits$. $\square$

Lemma 15.77.13. Let $(A_ n)$ be an inverse system of abelian groups. The following are equivalent

1. $(A_ n)$ is zero as a pro-object,

2. $\mathop{\mathrm{lim}}\nolimits A_ n = 0$ and $R^1\mathop{\mathrm{lim}}\nolimits A_ n = 0$ and the same holds for $\bigoplus _{i \in \mathbf{N}} (A_ n)$.

Proof. It follows from Lemma 15.77.10 that (1) implies (2). Assume (2). Then $(A_ n)$ is ML by Lemma 15.77.11. 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. $\square$

[1] To use these spectral sequences we have to show that $\textit{Ab}(\mathbf{N})$ has enough injectives. A inverse system $(I_ n)$ of abelian groups is injective if and only if each $I_ n$ is an injective abelian group and the transition maps are split surjections. Every system embeds in one of these. Details omitted.

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