Lemma 50.17.2. With notation as in More on Morphisms, Lemma 37.17.3 and denoting $f : X \to S$ the structure morphism there is a canonical distinguished triangle
\[ \Omega ^\bullet _{X/S} \to Rb_*(\Omega ^\bullet _{X'/S}) \oplus i_*\Omega ^\bullet _{Z/S} \to i_*Rp_*(\Omega ^\bullet _{E/S}) \to \Omega ^\bullet _{X/S}[1] \]
in $D(X, f^{-1}\mathcal{O}_ S)$ where the four maps
\[ \begin{matrix} \Omega ^\bullet _{X/S}
& \to
& Rb_*(\Omega ^\bullet _{X'/S}),
\\ \Omega ^\bullet _{X/S}
& \to
& i_*\Omega ^\bullet _{Z/S},
\\ Rb_*(\Omega ^\bullet _{X'/S})
& \to
& i_*Rp_*(\Omega ^\bullet _{E/S}),
\\ i_*\Omega ^\bullet _{Z/S}
& \to
& i_*Rp_*(\Omega ^\bullet _{E/S})
\end{matrix} \]
are the canonical ones (Section 50.2), except with sign reversed for one of them.
Proof.
Choose a distinguished triangle
\[ C \to Rb_*\Omega ^\bullet _{X'/S} \oplus i_*\Omega ^\bullet _{Z/S} \to i_*Rp_*\Omega ^\bullet _{E/S} \to C[1] \]
in $D(X, f^{-1}\mathcal{O}_ S)$. It suffices to show that $\Omega ^\bullet _{X/S}$ is isomorphic to $C$ in a manner compatible with the canonical maps. By the axioms of triangulated categories there exists a map of distinguished triangles
\[ \xymatrix{ C' \ar[r] \ar[d] & b_*\Omega ^\bullet _{X'/S} \oplus i_*\Omega ^\bullet _{Z/S} \ar[r] \ar[d] & i_*p_*\Omega ^\bullet _{E/S} \ar[r] \ar[d] & C'[1] \ar[d] \\ C \ar[r] & Rb_*\Omega ^\bullet _{X'/S} \oplus i_*\Omega ^\bullet _{Z/S} \ar[r] & i_*Rp_*\Omega ^\bullet _{E/S} \ar[r] & C[1] } \]
By Lemma 50.17.1 part (3) and Derived Categories, Proposition 13.4.23 we conclude that $C' \to C$ is an isomorphism. By Lemma 50.17.1 part (2) the map $i_*\Omega ^\bullet _{Z/S} \to i_*p_*\Omega ^\bullet _{E/S}$ is an isomorphism. Thus $C' = b_*\Omega ^\bullet _{X'/S}$ in the derived category. Finally we use Lemma 50.17.1 part (1) tells us this is equal to $\Omega ^\bullet _{X/S}$. We omit the verification this is compatible with the canonical maps.
$\square$
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