15.86 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 15.86.6. Many of the arguments in this section duplicate the arguments used to construct the cohomological machinery for sheaves of abelian groups on sites.
Lemma 15.86.1. The functor \mathop{\mathrm{lim}}\nolimits : \textit{Ab}(\mathbf{N}) \to \textit{Ab} has a right derived functor
15.86.1.1
\begin{equation} \label{more-algebra-equation-Rlim} R\mathop{\mathrm{lim}}\nolimits : D(\textit{Ab}(\mathbf{N})) \longrightarrow D(\textit{Ab}) \end{equation}
As usual we set R^ p\mathop{\mathrm{lim}}\nolimits (K) = H^ p(R\mathop{\mathrm{lim}}\nolimits (K)). Moreover, we have
for any (A_ n) in \textit{Ab}(\mathbf{N}) we have R^ p\mathop{\mathrm{lim}}\nolimits A_ n = 0 for p > 1,
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,
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 ,
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
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 12.31). 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
Lemma 15.86.2. Let
0 \to (A_ i) \to (B_ i) \to (C_ i) \to 0
be a short exact sequence of inverse systems of abelian groups. Then there is an associated 6 term exact sequence 0 \to \mathop{\mathrm{lim}}\nolimits A_ i \to \mathop{\mathrm{lim}}\nolimits B_ i \to \mathop{\mathrm{lim}}\nolimits C_ i \to R^1\mathop{\mathrm{lim}}\nolimits A_ i \to R^1\mathop{\mathrm{lim}}\nolimits B_ i \to R^1\mathop{\mathrm{lim}}\nolimits C_ i \to 0.
Proof.
Follows from the vanishing in Lemma 15.86.1.
\square
Here is the “correct” formulation of Homology, Lemma 12.31.7.
Lemma 15.86.3. Let
(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
(A^{-2}_ n) and (A^{-1}_ n) have vanishing R^1\mathop{\mathrm{lim}}\nolimits ,
(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 nth 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 15.86.4. 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 13.34.1 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
Lemma 15.86.5. 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 13.33.1 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
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 13.34.2.
Lemma 15.86.7. 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 13.34.1.
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
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 15.86.1 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
Lemma 15.86.8. With notation as in Lemma 15.86.7 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 15.86.1.
\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 15.86.9. 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
Lemma 15.86.11. 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 15.86.8 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
\prod B_ n \longrightarrow \prod B_ n
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
Lemma 15.86.12 (Emmanouil).reference Let (A_ n) be an inverse system of abelian groups. The following are equivalent
(A_ n) is Mittag-Leffler,
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 15.86.1.
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
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 ith 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
Lemma 15.86.13. Let
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 15.86.12, the fact that taking infinite direct sums is exact, and the long exact sequence of cohomology associated to R\mathop{\mathrm{lim}}\nolimits .
\square
Lemma 15.86.14. Let (A_ n) be an inverse system of abelian groups. The following are equivalent
(A_ n) is zero as a pro-object,
\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 15.86.11 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
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