First, let $\mathcal{I}$ be an injective abelian sheaf on $\mathcal{C}$. Then the spectral sequence of Lemma \ref{lemma-augmentation-spectral-sequence} for the sheaf $a^{-1}\mathcal{I}$ degenerates as $(a^{-1}\mathcal{I})_p = a_p^{-1}\mathcal{I}$ is injective by Lemma \ref{lemma-localize-injective}. Thus the complex $$ a_{0, *}a_0^{-1}\mathcal{I} \to a_{1, *}a_1^{-1}\mathcal{I} \to a_{2, *}a_2^{-1}\mathcal{I} \to\ldots $$ computes $Ra_*a^{-1}\mathcal{I}$. By Lemma \ref{lemma-hypercovering-cech-complex} this is equal to the complex $\SheafHom(s(\mathbf{Z}_{F(K)}^\#), \mathcal{I})$. Because $K$ is a hypercovering, we see that $s(\mathbf{Z}_{F(K)}^\#)$ is exact in degrees $> 0$ by Hypercoverings, Lemma \ref{hypercovering-lemma-acyclic-hypercover-sheaves} applied to the simplicial presheaf $F(K)$. Since $\mathcal{I}$ is injective, the functor $\SheafHom(-, \mathcal{I})$ is exact and we conclude that $\SheafHom(s(\mathbf{Z}_{F(K)}^\#), \mathcal{I})$ is exact in positive degrees. We conclude that $R^pa_*a^{-1}\mathcal{I} = 0$ for $p > 0$. On the other hand, we have $\mathcal{I} = a_*a^{-1}\mathcal{I}$ by Lemma \ref{lemma-hypercovering-descent-sheaves}. \medskip\noindent Bounded case. Let $E \in D^+(\mathcal{C})$. Choose a bounded below complex $\mathcal{I}^\bullet$ of injectives representing $E$. By the result of the first paragraph and Leray's acyclicity lemma (Derived Categories, Lemma \ref{derived-lemma-leray-acyclicity}) $Ra_*a^{-1}\mathcal{I}^\bullet$ is computed by the complex $a_*a^{-1}\mathcal{I}^\bullet = \mathcal{I}^\bullet$ and we conclude the lemma is true in this case. \medskip\noindent Unbounded case. We urge the reader to skip this, since the argument is the same as above, except that we use explicit representation by double complexes to get around convergence issues. Let $E \in D(\mathcal{C})$. To show the map $E \to Ra_*a^{-1}E$ is an isomorphism, it suffices to show for every object $U$ of $\mathcal{C}$ that $$ R\Gamma(U, E) = R\Gamma(U, Ra_*a^{-1}E) $$ We will compute both sides and show the map $E \to Ra_*a^{-1}E$ induces an isomorphism. Choose a K-injective complex $\mathcal{I}^\bullet$ representing $E$. Choose a quasi-isomorphism $a^{-1}\mathcal{I}^\bullet \to \mathcal{J}^\bullet$ for some K-injective complex $\mathcal{J}^\bullet$ on $(\mathcal{C}/K)_{total}$. We have $$ R\Gamma(U, E) = R\Hom(\mathbf{Z}_U^\#, E) $$ and $$ R\Gamma(U, Ra_*a^{-1}E) = R\Hom(\mathbf{Z}_U^\#, Ra_*a^{-1}E) = R\Hom(a^{-1}\mathbf{Z}_U^\#, a^{-1}E) $$ By Lemma \ref{lemma-simplicial-resolution-augmentation} we have a quasi-isomorphism $$ \Big(\ldots \to g_{2!}(a_2^{-1}\mathbf{Z}_U^\#) \to g_{1!}(a_1^{-1}\mathbf{Z}_U^\#) \to g_{0!}(a_0^{-1}\mathbf{Z}_U^\#)\Big) \longrightarrow a^{-1}\mathbf{Z}_U^\# $$ Hence $R\Hom(a^{-1}\mathbf{Z}_U^\#, a^{-1}E)$ is equal to $$ R\Gamma((\mathcal{C}/K)_{total}, R\SheafHom( \ldots \to g_{2!}(a_2^{-1}\mathbf{Z}_U^\#) \to g_{1!}(a_1^{-1}\mathbf{Z}_U^\#) \to g_{0!}(a_0^{-1}\mathbf{Z}_U^\#), \mathcal{J}^\bullet)) $$ By the construction in Cohomology on Sites, Section \ref{sites-cohomology-section-internal-hom} and since $\mathcal{J}^\bullet$ is K-injective, we see that this is represented by the complex of abelian groups with terms $$ \prod\nolimits_{p + q = n} \Hom(g_{p!}(a_p^{-1}\mathbf{Z}_U^\#), \mathcal{J}^q) = \prod\nolimits_{p + q = n} \Hom(a_p^{-1}\mathbf{Z}_U^\#, g_p^{-1}\mathcal{J}^q) $$ See Cohomology on Sites, Lemmas \ref{sites-cohomology-lemma-RHom-into-K-injective} and \ref{sites-cohomology-lemma-section-RHom-over-U} for more information. Thus we find that $R\Gamma(U, Ra_*a^{-1}E)$ is computed by the product total complex $\text{Tot}_\pi(B^{\bullet, \bullet})$ with $B^{p, q} = \Hom(a_p^{-1}\mathbf{Z}_U^\#, g_p^{-1}\mathcal{J}^q)$. For the other side we argue similarly. First we note that $$ s(\mathbf{Z}_{F(K)}^\#) \longrightarrow \mathbf{Z} $$ is a quasi-isomorphism of complexes on $\mathcal{C}$ by Hypercoverings, Lemma \ref{hypercovering-lemma-acyclic-hypercover-sheaves}. Since $\mathbf{Z}_U^\#$ is a flat sheaf of $\mathbf{Z}$-modules we see that $$ s(\mathbf{Z}_{F(K)}^\#) \otimes_\mathbf{Z} \mathbf{Z}_U^\# \longrightarrow \mathbf{Z}_U^\# $$ is a quasi-isomorphism. Therefore $R\Hom(\mathbf{Z}_U^\#, E)$ is equal to $$ R\Gamma(\mathcal{C}, R\SheafHom( s(\mathbf{Z}_{F(K)}^\#) \otimes_\mathbf{Z} \mathbf{Z}_U^\#, \mathcal{I}^\bullet)) $$ By the construction of $R\SheafHom$ and since $\mathcal{I}^\bullet$ is K-injective, this is represented by the complex of abelian groups with terms $$ \prod\nolimits_{p + q = n} \Hom(\mathbf{Z}^\#_{K_p} \otimes_\mathbf{Z} \mathbf{Z}_U^\#, \mathcal{I}^q) = \prod\nolimits_{p + q = n} \Hom(a_p^{-1}\mathbf{Z}_U^\#, a_p^{-1}\mathcal{I}^q) $$ The equality of terms follows from the fact that $\mathbf{Z}^\#_{K_p} \otimes_\mathbf{Z} \mathbf{Z}_U^\# = a_{p!}a_p^{-1}\mathbf{Z}_U^\#$ by Modules on Sites, Remark \ref{sites-modules-remark-j-shriek-tensor}. Thus we find that $R\Gamma(U, E)$ is computed by the product total complex $\text{Tot}_\pi(A^{\bullet, \bullet})$ with $A^{p, q} = \Hom(a_p^{-1}\mathbf{Z}_U^\#, a_p^{-1}\mathcal{I}^q)$. \medskip\noindent Since $\mathcal{I}^\bullet$ is K-injective we see that $a_p^{-1}\mathcal{I}^\bullet$ is K-injective, see Lemma \ref{lemma-localize-injective}. Since $\mathcal{J}^\bullet$ is K-injective we see that $g_p^{-1}\mathcal{J}^\bullet$ is K-injective, see Lemma \ref{lemma-restriction-injective-to-component-site}. Both represent the object $a_p^{-1}E$. Hence for every $p \geq 0$ the map of complexes $$ A^{p, \bullet} = \Hom(a_p^{-1}\mathbf{Z}_U^\#, a_p^{-1}\mathcal{I}^\bullet) \longrightarrow \Hom(a_p^{-1}\mathbf{Z}_U^\#, g_p^{-1}\mathcal{J}^\bullet) = B^{p, \bullet} $$ induced by $g_p^{-1}$ applied to the given map $a^{-1}\mathcal{I}^\bullet \to \mathcal{J}^\bullet$ is a quasi-isomorphisms as these complexes both compute $$ R\Hom(a_p^{-1}\mathbf{Z}_U^\#, a_p^{-1}E) $$ By More on Algebra, Lemma \ref{more-algebra-lemma-prod-qis-gives-qis} we conclude that the right vertical arrow in the commutative diagram $$ \xymatrix{ R\Gamma(U, E) \ar[r] \ar[d] & \text{Tot}_\pi(A^{\bullet, \bullet}) \ar[d] \\ R\Gamma(U, Ra_*a^{-1}E) \ar[r] & \text{Tot}_\pi(B^{\bullet, \bullet}) } $$ is a quasi-isomorphism. Since we saw above that the horizontal arrows are quasi-isomorphisms, so is the left vertical arrow.