Lemma 15.93.6. Let $A$ be a ring and let $f \in A$. Let $K$ be an object of $D(A)$. Denote $K_ n = K \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A)$. For all $p \in \mathbf{Z}$ there is a commutative diagram
\[ \xymatrix{ & 0 & 0 \\ 0 \ar[r] & \widehat{H^ p(K)} \ar[r] \ar[u] & \mathop{\mathrm{lim}}\nolimits H^ p(K_ n) \ar[r] \ar[u] & T_ f(H^{p + 1}(K)) \ar[r] & 0 \\ 0 \ar[r] & H^0(H^ p(K)^\wedge ) \ar[r] \ar[u] & H^ p(K^\wedge ) \ar[r] \ar[u] & T_ f(H^{p + 1}(K)) \ar[r] \ar@{=}[u] & 0 \\ & R^1\mathop{\mathrm{lim}}\nolimits H^ p(K)[f^ n] \ar[u] \ar[r]^\cong & R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(K_ n) \ar[u] \\ & 0 \ar[u] & 0 \ar[u] } \]
with exact rows and columns where $\widehat{H^ p(K)} = \mathop{\mathrm{lim}}\nolimits H^ p(K)/f^ nH^ p(K)$ is the usual $f$-adic completion. The left vertical short exact sequence and the middle horizontal short exact sequence are taken from Example 15.93.5 The middle vertical short exact sequence is the one from Lemma 15.87.4.
Proof.
To construct the top horizontal short exact sequence, observe that we have the following inverse system short exact sequences
\[ 0 \to H^ p(K)/f^ nH^ p(K) \to H^ p(K_ n) \to H^{p + 1}(K)[f^ n] \to 0 \]
coming from the construction of $K_ n$ as a shift of the cone on $f^ n : K \to K$. Taking the inverse limit of these we obtain the top horizontal short exact sequence, see Homology, Lemma 12.31.3.
Let us prove that we have a commutative diagram as in the lemma. We consider the map $L = \tau _{\leq p}K \to K$. Setting $L_ n = L \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A)$ we obtain a map $(L_ n) \to (K_ n)$ of inverse systems which induces a map of short exact sequences
\[ \xymatrix{ 0 & 0 \\ \mathop{\mathrm{lim}}\nolimits H^ p(L_ n) \ar[r] \ar[u] & \mathop{\mathrm{lim}}\nolimits H^ p(K_ n) \ar[u] \\ H^ p(L^\wedge ) \ar[r] \ar[u] & H^ p(K^\wedge ) \ar[u] \\ R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(L_ n) \ar[r] \ar[u] & R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(K_ n) \ar[u] \\ 0 \ar[u] & 0 \ar[u] } \]
Since $H^ i(L) = 0$ for $i > p$ and $H^ p(L) = H^ p(K)$, a computation using the references in the statement of the lemma shows that $H^ p(L^\wedge ) = H^0(H^ p(K)^\wedge )$ and that $H^ p(L_ n) = H^ p(K)/f^ nH^ p(K)$. On the other hand, we have $H^{p - 1}(L_ n) = H^{p - 1}(K_ n)$ and hence we see that we get the isomorphism as indicated in the statement of the lemma since we already know the kernel of $H^0(H^ p(K)^\wedge ) \to \widehat{H^ p(K)}$ is equal to $R^1\mathop{\mathrm{lim}}\nolimits H^ p(K)[f^ n]$. We omit the verification that the rightmost square in the diagram commutes if we define the top row by the construction in the first paragraph of the proof.
$\square$
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