Show that \(\varprojlim^1 G_i = 0\) if the sequence \(\cdots \to G_2 \xrightarrow{\alpha_2} G_1 \xrightarrow{\alpha_1} G_0\) satisfies the Mittag–Leffler condition that for each \(i\) the images of the maps \(G_{i+n} \to G_i\) are independent of \(n\) for sufficiently large \(n\).
Solution
For a tower \[ \cdots\xrightarrow{\alpha_2}G_1\xrightarrow{\alpha_1}G_0, \] write \[ \Phi:\prod_{i\ge0}G_i\longrightarrow\prod_{i\ge0}G_i, \qquad \Phi((g_i))=(g_i-\alpha_{i+1}(g_{i+1}))_i. \] By definition, \[ \varprojlim G_i=\ker\Phi, \qquad \varprojlim{}^1G_i=\operatorname{coker}\Phi. \] Thus it suffices to prove that \(\Phi\) is surjective under the Mittag–Leffler hypothesis.
For each \(i\), let \[ I_i=\operatorname{im}(G_j\to G_i) \] for all sufficiently large \(j\); this is well-defined by Mittag–Leffler stabilization. The bonding maps carry \(I_{i+1}\) onto \(I_i\). Indeed, choose \(j\) large enough that both stable images have been reached. Then \[ I_i=\operatorname{im}(G_j\to G_i) =\alpha_{i+1}\bigl(\operatorname{im}(G_j\to G_{i+1})\bigr) =\alpha_{i+1}(I_{i+1}). \] Hence the tower \((I_i)\) has surjective bonding maps, so its \(\varprojlim{}^1\) vanishes: given \((b_i)\in\prod I_i\), choose \(g_0\) arbitrarily and then recursively choose \(g_{i+1}\) with \[ \alpha_{i+1}(g_{i+1})=g_i-b_i. \] This solves \(\Phi(g)=b\).
Now let \(Q_i=G_i/I_i\). The quotient tower is pro-zero: for every \(i\) some sufficiently long composite \[ Q_j\longrightarrow Q_i \] is zero, since the image of \(G_j\to G_i\) is then exactly \(I_i\). Passing to a cofinal subsequence, we may assume each bonding map \(Q_{i+1}\to Q_i\) is zero. For a zero-bonding tower, \[ \Phi((q_i))=(q_i), \] so \(\Phi\) is an isomorphism and \(\varprojlim{}^1Q_i=0\).
The short exact sequence of towers \[ 0\to I_i\to G_i\to Q_i\to0 \] gives the standard exact sequence for inverse limits and their first derived limits. Since both outer \(\varprojlim{}^1\) terms vanish, so does the middle one: \[ \boxed{\varprojlim{}^1G_i=0.} \]