Lemma 21.29.2. With $\epsilon : (\mathcal{C}_\tau , \mathcal{O}_\tau ) \to (\mathcal{C}_{\tau '}, \mathcal{O}_{\tau '})$ as above. Let $A$ be a set and for $\alpha \in A$ let
\[ \xymatrix{ E_\alpha \ar[d] \ar[r] & Y_\alpha \ar[d] \\ Z_\alpha \ar[r] & X_\alpha } \]
be a commutative diagram in the category $\mathcal{C}$. Assume that
a $\tau '$-sheaf $\mathcal{F}'$ is a $\tau $-sheaf if $\mathcal{F}'(X_\alpha ) = \mathcal{F}'(Z_\alpha ) \times _{\mathcal{F}'(E_\alpha )} \mathcal{F}'(Y_\alpha )$ for all $\alpha $,
for $K'$ in $D(\mathcal{O}_{\tau '})$ in the essential image of $R\epsilon _*$ the maps $c^{K'}_{X_\alpha , Z_\alpha , Y_\alpha , E_\alpha }$ of Lemma 21.26.1 are isomorphisms for all $\alpha $.
Then $K' \in D^+(\mathcal{O}_{\tau '})$ is in the essential image of $R\epsilon _*$ if and only if the maps $c^{K'}_{X_\alpha , Z_\alpha , Y_\alpha , E_\alpha }$ are isomorphisms for all $\alpha $.
Proof.
The “only if” direction is implied by assumption (2). On the other hand, if $K'$ has a unique nonzero cohomology sheaf, then the “if” direction follows from assumption (1). In general we will use an induction argument to prove the “if” direction. Let us say an object $K'$ of $D^+(\mathcal{O}_{\tau '})$ satisfies (P) if the maps $c^{K'}_{X_\alpha , Z_\alpha , Y_\alpha , E_\alpha }$ are isomorphisms for all $\alpha \in A$.
Namely, let $K'$ be an object of $D^+(\mathcal{O}_{\tau '})$ satisfying (P). Choose a distinguished triangle
\[ K' \to R\epsilon _*\epsilon ^{-1}K' \to M' \to K'[1] \]
in $D^+(\mathcal{O}_{\tau '})$ where the first arrow is the adjuntion map. By (2) and Lemma 21.26.2 we see that $M'$ has (P). On the other hand, applying $\epsilon ^{-1}$ and using that $\epsilon ^{-1}R\epsilon _* = \text{id}$ by Section 21.27 we find that $\epsilon ^{-1}M' = 0$. In the next paragraph we will show $M' = 0$ which finishes the proof.
Let $K'$ be an object of $D^+(\mathcal{O}_{\tau '})$ satisfying (P) with $\epsilon ^{-1}K' = 0$. We will show $K' = 0$. Namely, given $n \in \mathbf{Z}$ such that $H^ i(K') = 0$ for $i < n$ we will show that $H^ n(K') = 0$. For $\alpha \in A$ we have a distinguished triangle
\[ R\Gamma _{\tau '}(X_\alpha , K') \to R\Gamma _{\tau '}(Z_\alpha , K') \oplus R\Gamma _{\tau '}(Y_\alpha , K') \to R\Gamma _{\tau '}(E_\alpha , K') \to R\Gamma _{\tau '}(X_\alpha , K')[1] \]
by Lemma 21.26.1. Taking cohomology in degree $n$ and using the assumed vanishing of cohomology sheaves of $K'$ we obtain an exact sequence
\[ 0 \to H^ n_{\tau '}(X_\alpha , K') \to H^ n_{\tau '}(Z_\alpha , K') \oplus H^ n_{\tau '}(Y_\alpha , K') \to H^ n_{\tau '}(E_\alpha , K') \]
which is the same as the exact sequence
\[ 0 \to \Gamma (X_\alpha , H^ n(K')) \to \Gamma (Z_\alpha , H^ n(K')) \oplus \Gamma (Y_\alpha , H^ n(K')) \to \Gamma (E_\alpha , H^ n(K')) \]
We conclude that $H^ n(K')$ is a a $\tau $-sheaf by assumption (1). However, since the $\tau $-sheafification $\epsilon ^{-1}H^ n(K') = H^ n(\epsilon ^{-1}K')$ is $0$ as $\epsilon ^{-1}K' = 0$ we conclude that $H^ n(K') = 0$ as desired.
$\square$
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