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

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. $\square$

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