The Stacks project

Proof. Omitted. $\square$

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

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

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. $\square$

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. $\square$

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

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. $\square$

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

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

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

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. $\square$

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. $\square$