The Stacks project

12.31 Inverse systems

Let $\mathcal{C}$ be a category. In Categories, Section 4.21 we defined the notion of an inverse system over a preordered set (with values in the category $\mathcal{C}$). If the preordered set is $\mathbf{N} = \{ 1, 2, 3, \ldots \} $ with the usual ordering such an inverse system over $\mathbf{N}$ is often simply called an inverse system. It consists quite simply of a pair $(M_ i, f_{ii'})$ where each $M_ i$, $i \in \mathbf{N}$ is an object of $\mathcal{C}$, and for each $i > i'$, $i, i' \in \mathbf{N}$ a morphism $f_{ii'} : M_ i \to M_{i'}$ such that moreover $f_{i'i''} \circ f_{ii'} = f_{ii''}$ whenever this makes sense. It is clear that in fact it suffices to give the morphisms $M_2 \to M_1$, $M_3 \to M_2$, and so on. Hence an inverse system is frequently pictured as follows

\[ M_1 \xleftarrow {\varphi _2} M_2 \xleftarrow {\varphi _3} M_3 \leftarrow \ldots \]

Moreover, we often omit the transition maps $\varphi _ i$ from the notation and we simply say “let $(M_ i)$ be an inverse system”.

The collection of all inverse systems with values in $\mathcal{C}$ forms a category with the obvious notion of morphism.

Lemma 12.31.1. Let $\mathcal{C}$ be a category.

  1. If $\mathcal{C}$ is an additive category, then the category of inverse systems with values in $\mathcal{C}$ is an additive category.

  2. If $\mathcal{C}$ is an abelian category, then the category of inverse systems with values in $\mathcal{C}$ is an abelian category. A sequence $(K_ i) \to (L_ i) \to (M_ i)$ of inverse systems is exact if and only if each $K_ i \to L_ i \to N_ i$ is exact.

Proof. Omitted. $\square$

The limit (see Categories, Section 4.21) of such an inverse system is denoted $\mathop{\mathrm{lim}}\nolimits M_ i$, or $\mathop{\mathrm{lim}}\nolimits _ i M_ i$. If $\mathcal{C}$ is the category of abelian groups (or sets), then the limit always exists and in fact can be described as follows

\[ \mathop{\mathrm{lim}}\nolimits _ i M_ i = \{ (x_ i) \in \prod M_ i \mid \varphi _ i(x_ i) = x_{i - 1}, \ i = 2, 3, \ldots \} \]

see Categories, Section 4.15. However, given a short exact sequence

\[ 0 \to (A_ i) \to (B_ i) \to (C_ i) \to 0 \]

of inverse systems of abelian groups it is not always the case that the associated system of limits is exact. In order to discuss this further we introduce the following notion.

Definition 12.31.2. Let $\mathcal{C}$ be an abelian category. We say the inverse system $(A_ i)$ satisfies the Mittag-Leffler condition, or for short is ML, if for every $i$ there exists a $c = c(i) \geq i$ such that

\[ \mathop{\mathrm{Im}}(A_ k \to A_ i) = \mathop{\mathrm{Im}}(A_ c \to A_ i) \]

for all $k \geq c$.

It turns out that the Mittag-Leffler condition is good enough to ensure that the $\mathop{\mathrm{lim}}\nolimits $-functor is exact, provided one works within the abelian category of abelian groups, modules over a ring, etc. It is shown in a paper by A. Neeman (see [Neeman-Counterexample]) that this condition is not strong enough in an abelian category having AB4* (having exact products).

Lemma 12.31.3. Let

\[ 0 \to (A_ i) \to (B_ i) \to (C_ i) \to 0 \]

be a short exact sequence of inverse systems of abelian groups.

  1. In any case the sequence

    \[ 0 \to \mathop{\mathrm{lim}}\nolimits _ i A_ i \to \mathop{\mathrm{lim}}\nolimits _ i B_ i \to \mathop{\mathrm{lim}}\nolimits _ i C_ i \]

    is exact.

  2. If $(B_ i)$ is ML, then also $(C_ i)$ is ML.

  3. If $(A_ i)$ is ML, then

    \[ 0 \to \mathop{\mathrm{lim}}\nolimits _ i A_ i \to \mathop{\mathrm{lim}}\nolimits _ i B_ i \to \mathop{\mathrm{lim}}\nolimits _ i C_ i \to 0 \]

    is exact.

Proof. Nice exercise. See Algebra, Lemma 10.87.1 for part (3). $\square$

Lemma 12.31.4. Let

\[ (A_ i) \to (B_ i) \to (C_ i) \to (D_ i) \]

be an exact sequence of inverse systems of abelian groups. If the system $(A_ i)$ is ML, then the sequence

\[ \mathop{\mathrm{lim}}\nolimits _ i B_ i \to \mathop{\mathrm{lim}}\nolimits _ i C_ i \to \mathop{\mathrm{lim}}\nolimits _ i D_ i \]

is exact.

Proof. Let $Z_ i = \mathop{\mathrm{Ker}}(C_ i \to D_ i)$ and $I_ i = \mathop{\mathrm{Im}}(A_ i \to B_ i)$. Then $\mathop{\mathrm{lim}}\nolimits Z_ i = \mathop{\mathrm{Ker}}(\mathop{\mathrm{lim}}\nolimits C_ i \to \mathop{\mathrm{lim}}\nolimits D_ i)$ and we get a short exact sequence of systems

\[ 0 \to (I_ i) \to (B_ i) \to (Z_ i) \to 0 \]

Moreover, by Lemma 12.31.3 we see that $(I_ i)$ has (ML), thus another application of Lemma 12.31.3 shows that $\mathop{\mathrm{lim}}\nolimits B_ i \to \mathop{\mathrm{lim}}\nolimits Z_ i$ is surjective which proves the lemma. $\square$

The following characterization of essentially constant inverse systems shows in particular that they have ML.

Lemma 12.31.5. Let $\mathcal{A}$ be an abelian category. Let $(A_ i)$ be an inverse system in $\mathcal{A}$ with limit $A = \mathop{\mathrm{lim}}\nolimits A_ i$. Then $(A_ i)$ is essentially constant (see Categories, Definition 4.22.1) if and only if there exists an $i$ and for all $j \geq i$ a direct sum decomposition $A_ j = A \oplus Z_ j$ such that (a) the maps $A_{j'} \to A_ j$ are compatible with the direct sum decompositions, (b) for all $j$ there exists some $j' \geq j$ such that $Z_{j'} \to Z_ j$ is zero.

Proof. Assume $(A_ i)$ is essentially constant. Then there exists an $i$ and a morphism $A_ i \to A$ such that $A \to A_ i \to A$ is the identity and for all $j \geq i$ there exists a $j' \geq j$ such that $A_{j'} \to A_ j$ factors as $A_{j'} \to A_ i \to A \to A_ j$ (the last map comes from $A = \mathop{\mathrm{lim}}\nolimits A_ i$). Hence setting $Z_ j = \mathop{\mathrm{Ker}}(A_ j \to A)$ for all $j \geq i$ works. Proof of the converse is omitted. $\square$

We will improve on the following lemma in More on Algebra, Lemma 15.86.13.

Lemma 12.31.6. Let

\[ 0 \to (A_ i) \to (B_ i) \to (C_ i) \to 0 \]

be an exact sequence of inverse systems of abelian groups. If $(C_ i)$ is essentially constant, then $(A_ i)$ has ML if and only if $(B_ i)$ has ML.

Proof. After renumbering we may assume that $C_ i = C \oplus Z_ i$ compatible with transition maps and that for all $i$ there exists an $i' \geq i$ such that $Z_{i'} \to Z_ i$ is zero, see Lemma 12.31.5.

First, assume $C = 0$, i.e., we have $C_ i = Z_ i$. In this case choose $1 = n_1 < n_2 < n_3 < \ldots $ such that $Z_{n_{i + 1}} \to Z_{n_ i}$ is zero. Then $B_{n_{i + 1}} \to B_{n_ i}$ factors through $A_{n_ i} \subset B_{n_ i}$. It follows that for $j \geq i + 1$ we have

\[ \mathop{\mathrm{Im}}(A_{n_ j} \to A_{n_ i}) \subset \mathop{\mathrm{Im}}(B_{n_ j} \to B_{n_ i}) \subset \mathop{\mathrm{Im}}(A_{n_{j - 1}} \to A_{n_ i}) \]

as subsets of $A_{n_ i}$. Thus the images $\mathop{\mathrm{Im}}(A_{n_ j} \to A_{n_ i})$ stabilize for $j \geq i + 1$ if and only if the same is true for the images $\mathop{\mathrm{Im}}(B_{n_ j} \to B_{n_ i})$. The equivalence follows from this (small detail omitted).

If $C \not= 0$, denote $B'_ i \subset B_ i$ the inverse image of $C$ by the map $B_ i \to C \oplus Z_ i$. Then by the previous paragraph we see that $(B'_ i)$ has ML if and only if $(B_ i)$ has ML. Thus we may replace $(B_ i)$ by $(B'_ i)$. In this case we have exact sequences $0 \to A_ i \to B_ i \to C \to 0$ for all $i$. It follows that $0 \to \mathop{\mathrm{Im}}(A_ j \to A_ i) \to \mathop{\mathrm{Im}}(B_ j \to B_ i) \to C \to 0$ is short exact for all $j \geq i$. Hence the images $\mathop{\mathrm{Im}}(A_ j \to A_ i)$ stabilize for $j \geq i$ if and only if the same is true for $\mathop{\mathrm{Im}}(B_ j \to B_ i)$ as desired. $\square$

The “correct” version of the following lemma is More on Algebra, Lemma 15.86.3.

Lemma 12.31.7. Let

\[ (A^{-2}_ i \to A^{-1}_ i \to A^0_ i \to A^1_ i) \]

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 $(H_ i^{-1})$, $(H_ i^0)$ 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 $(A^{-2}_ i)$ and $(A^{-1}_ i)$ are ML and $(H^{-1}_ i)$ is essentially constant, then $H^0 = \mathop{\mathrm{lim}}\nolimits H_ i^0$.

Proof. Let $Z^ j_ i = \mathop{\mathrm{Ker}}(A^ j_ i \to A^{j + 1}_ i)$ and $I^ j_ i = \mathop{\mathrm{Im}}(A^{j - 1}_ i \to A^ j_ i)$. Note that $\mathop{\mathrm{lim}}\nolimits Z^0_ i = \mathop{\mathrm{Ker}}(\mathop{\mathrm{lim}}\nolimits A^0_ i \to \mathop{\mathrm{lim}}\nolimits A^1_ i)$ as taking kernels commutes with limits. The systems $(I^{-1}_ i)$ and $(I^0_ i)$ have ML as quotients of the systems $(A^{-2}_ i)$ and $(A^{-1}_ i)$, see Lemma 12.31.3. Thus an exact sequence

\[ 0 \to (I^{-1}_ i) \to (Z^{-1}_ i) \to (H^{-1}_ i) \to 0 \]

of inverse systems where $(I^{-1}_ i)$ has ML and where $(H^{-1}_ i)$ is essentially constant by assumption. Hence $(Z^{-1}_ i)$ has ML by Lemma 12.31.6. The exact sequence

\[ 0 \to (Z^{-1}_ i) \to (A^{-1}_ i) \to (I^0_ i) \to 0 \]

and an application of Lemma 12.31.3 shows that $\mathop{\mathrm{lim}}\nolimits A^{-1}_ i \to \mathop{\mathrm{lim}}\nolimits I^0_ i$ is surjective. Finally, the exact sequence

\[ 0 \to (I^0_ i) \to (Z^0_ i) \to (H^0_ i) \to 0 \]

and Lemma 12.31.3 show that $\mathop{\mathrm{lim}}\nolimits I^0_ i \to \mathop{\mathrm{lim}}\nolimits Z^0_ i \to \mathop{\mathrm{lim}}\nolimits H^0_ i \to 0$ is exact. Putting everything together we win. $\square$

Sometimes we need a version of the lemma above where we take limits over big ordinals.

Lemma 12.31.8. Let $\alpha $ be an ordinal. Let $K_\beta ^\bullet $, $\beta < \alpha $ be an inverse system of complexes of abelian groups over $\alpha $. If for all $\beta < \alpha $ the complex $K_\beta ^\bullet $ is acyclic and the map

\[ K^ n_\beta \longrightarrow \mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K^ n_\gamma \]

is surjective, then the complex $\mathop{\mathrm{lim}}\nolimits _{\beta < \alpha } K_\beta ^\bullet $ is acyclic.

Proof. By transfinite induction we prove this holds for every ordinal $\alpha $ and every system as in the lemma. In particular, whilst proving the result for $\alpha $ we may assume the complexes $\mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K^ n_\gamma $ are acyclic.

Let $x \in \mathop{\mathrm{lim}}\nolimits _{\beta < \alpha } K^0_\alpha $ with $\text{d}(x) = 0$. We will find a $y \in K^{-1}_\alpha $ with $\text{d}(y) = x$. Write $x = (x_\beta )$ where $x_\beta \in K_\beta ^0$ is the image of $x$ for $\beta < \alpha $. We will construct $y = (y_\beta )$ by transfinite recursion.

For $\beta = 0$ let $y_0 \in K_0^{-1}$ be any element with $\text{d}(y_0) = x_0$.

For $\beta = \gamma + 1$ a successor, we have to find an element $y_\beta $ which maps both to $y_\gamma $ by the transition map $f : K^\bullet _\beta \to K^\bullet _\gamma $ and to $x_\beta $ under the differential. As a first approximation we choose $y'_\beta $ with $\text{d}(y'_\beta ) = x_\beta $. Then the difference $y_\gamma - f(y'_\beta )$ is in the kernel of the differential, hence equal to $\text{d}(z_\gamma )$ for some $z_\gamma \in K^{-2}_\gamma $. By assumption, the map $f^{-2} : K^{-2}_\beta \to K^{-2}_\gamma $ is surjective. Hence we write $z_\gamma = f(z_\beta )$ and change $y'_\beta $ into $y_\beta = y'_\beta + \text{d}(z_\beta )$ which works.

If $\beta $ is a limit ordinal, then we have the element $(y_\gamma )_{\gamma < \beta }$ in $\mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K^{-1}_\gamma $ whose differential is the image of $x_\beta $. Thus we can argue in exactly the same manner as above using the termwise surjective map of complexes $f : K_\beta ^\bullet \to \mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K_\gamma ^\bullet $ and the fact (see first paragraph of proof) that we may assume $\mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K_\gamma ^\bullet $ is acyclic by induction. $\square$

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