Lemma 15.86.12 (Emmanouil). 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 $\bigoplus _{i \in \mathbf{N}} (A_ n)$.

Taken from [Emmanouil].

Lemma 15.86.12 (Emmanouil). 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 $i$th summand. We claim that $\xi $ is not in the image of the map $\prod B_ n \to \prod B_ n$ of Lemma 15.86.1. 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. $\square$

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