Lemma 47.12.3. With notation as in Lemma 47.12.1. For objects $K, L$ in $D(A)$ there is a canonical isomorphism

$R\mathop{\mathrm{Hom}}\nolimits _ A(K^\wedge , L^\wedge ) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(R\Gamma _ Z(K), R\Gamma _ Z(L))$

in $D(A)$.

Proof. Say $I = (f_1, \ldots , f_ r)$. Denote $C = (A \to \prod A_{f_ i} \to \ldots \to A_{f_1 \ldots f_ r})$ the alternating Čech complex. Then derived completion is given by $R\mathop{\mathrm{Hom}}\nolimits _ A(C, -)$ (More on Algebra, Lemma 15.91.10) and local cohomology by $C \otimes ^\mathbf {L} -$ (Lemma 47.9.1). Combining the isomorphism

$R\mathop{\mathrm{Hom}}\nolimits _ A(K \otimes ^\mathbf {L} C, L \otimes ^\mathbf {L} C) = R\mathop{\mathrm{Hom}}\nolimits _ A(K, R\mathop{\mathrm{Hom}}\nolimits _ A(C, L \otimes ^\mathbf {L} C))$

(More on Algebra, Lemma 15.73.1) and the map

$L \to R\mathop{\mathrm{Hom}}\nolimits _ A(C, L \otimes ^\mathbf {L} C)$

(More on Algebra, Lemma 15.73.6) we obtain a map

$\gamma : R\mathop{\mathrm{Hom}}\nolimits _ A(K, L) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(K \otimes ^\mathbf {L} C, L \otimes ^\mathbf {L} C)$

On the other hand, the right hand side is derived complete as it is equal to

$R\mathop{\mathrm{Hom}}\nolimits _ A(C, R\mathop{\mathrm{Hom}}\nolimits _ A(K, L \otimes ^\mathbf {L} C)).$

Thus $\gamma$ factors through the derived completion of $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ by the universal property of derived completion. However, the derived completion goes inside the $R\mathop{\mathrm{Hom}}\nolimits _ A$ by More on Algebra, Lemma 15.91.13 and we obtain the desired map.

To show that the map of the lemma is an isomorphism we may assume that $K$ and $L$ are derived complete, i.e., $K = K^\wedge$ and $L = L^\wedge$. In this case we are looking at the map

$\gamma : R\mathop{\mathrm{Hom}}\nolimits _ A(K, L) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(R\Gamma _ Z(K), R\Gamma _ Z(L))$

By Proposition 47.12.2 we know that the cohomology groups of the left and the right hand side coincide. In other words, we have to check that the map $\gamma$ sends a morphism $\alpha : K \to L$ in $D(A)$ to the morphism $R\Gamma _ Z(\alpha ) : R\Gamma _ Z(K) \to R\Gamma _ Z(L)$. We omit the verification (hint: note that $R\Gamma _ Z(\alpha )$ is just the map $\alpha \otimes \text{id}_ C : K \otimes ^\mathbf {L} C \to L \otimes ^\mathbf {L} C$ which is almost the same as the construction of the map in More on Algebra, Lemma 15.73.6). $\square$

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