no irreducible component of ${D_Y}_I$.

In the first case we can use Lemma \[lem:tracemaps\] with $Y=V$ and $Z=X=U$ to deduce the isomorphism

$$i_*\Omega_U\xrightarrow{\simeq} R\Gamma_{U}\Omega_V[2]\hspace{0.5cm}\text{in }D_+(Sh(V)).$$

In the second case we have a closed immersion $\emptyset\hookrightarrow V$ and using Lemma \[lem:tracemaps\] we find an isomorphism between trivial complexes. If we repeat the same reasoning for all irreducible components of ${D_X}_I$ we obtain an isomorphism

$$i_*\Omega_{{D_Y}_I}\xrightarrow{\simeq} R\Gamma_{{D_Y}_I}\Omega_{{D_X}_I}^{{\bullet}}[2]\hspace{0.5cm}\text{in }D_+(Sh({D_X}_I)).$$

where $i$ is the closed immersion ${D_Y}_I\hookrightarrow {D_X}_I$. If we do the same for all $I\in{\square}^*_{m_1}$ and then switch to semisimplicial objects, we obtain the isomorphism

$$\label{eq:isosheaves}
i_*\Omega_{{D_Y}_{{\bullet}}}^{{\bullet}}\xrightarrow{\simeq} R\Gamma_{{D_Y}_{{\bullet}}}\Omega_{{D_X}_{{\bullet}}}^{{\bullet}}[2]\hspace{0.5cm}\text{in }D_+(Sh({D_X}_{{\bullet}})).$$

where $i$ is the closed immersion ${D_Y}_{{\bullet}}\hookrightarrow{D_X}_{{\bullet}}$.

Denote by $j$ the closed immersion $D_Y\hookrightarrow D_X$ and by ${\varepsilon}_Y$ and ${\varepsilon}_X$ the natural augmentations ${D_Y}_{{\bullet}}\rightarrow D_Y$ and ${D_X}_{{\bullet}}\rightarrow D_X$; we have a commutative diagram

$$\xymatrixcolsep{1.5cm}\xymatrixrowsep{1.5cm}\xymatrix{
Sh({D_Y}_{{\bullet}}) \ar[r]^{i_*} \ar[d]^{{{\varepsilon}_Y}_*} & Sh({D_X}_{{\bullet}}) \ar[d]^{{{\varepsilon}_X}_*}\\
Sh(D_Y) \ar[r]^{j_*} & Sh(D_X).}$$ From this we deduce the equality of the total derived functors $R(j_*\circ{{\varepsilon}_Y}_*)=R({{\varepsilon}_X}_*\circ i_*)$. Both $i_*$ and $j_*$ are exact, because they are pushforwards of closed immersions, so they coincide with their derived functors; moreover, all pushforwards preserve injective objects, which are adapted to any functor. We thus obtain an isomorphism

$$\label{eq:comm2}
R{{\varepsilon}_X}_*\circ i_*=R{{\varepsilon}_X}_*\circ Ri_*\simeq R({{\varepsilon}_X}_*\circ i_*)=R(j_*\circ {{\varepsilon}_Y}_*)\simeq Rj_*\circ R{{\varepsilon}_Y}_*=j_*\circ R{{\varepsilon}_Y}_*$$

