# Detecting zero morphisms via an open subscheme and its complement.

In the setting described in Bernstein, Beilinson, and Deligne, associated to a scheme $X$, a closed subscheme $i: Z \to X$ and its open complement $j: U \to X$ we have six functors between the corresponding derived categories of étale sheaves $i^*, i_*, i^!, j_!, j^*, j_*$ (they are all derived but I will ommit the L's and R's).

My question is: if $X$ is a variety over a field $k$ with structural morphism $f: X \to k$ and $F$ is an object of $D^b(k)$, then is the canonical morphism $f^*F \to i_*i^*f^*F \oplus j_*j^*f^*F$ ever a monomorphism in the triangulated category $D^b(X)$?

-
What is $a^*$? . –  Piotr Achinger Mar 10 '12 at 5:29
My guess is $a=f$ –  Rami Mar 10 '12 at 5:35
Isn't it true that $j^*$, $i^*$, $a^*$ and $j_*$ are exact? Also $a^* F$ is flat, so $i^* a^* F$ is just the usual (non-derived) pull-back? Also, $F$ is a direct sum of its cohomology. That is, you are basically asking if $\mathcal{O}_X\to \mathcal{O}_Z\oplus\mathcal{O}_U$ is a monomorphism. Is that right? –  Piotr Achinger Mar 10 '12 at 5:46
Since sheaves on $X$ are a full subcategory of $D^b(X)$, if this is a monomorphism, then it is still a monomorphism in the category of sheaves on $X$. This I think may fail if $X$ has an embedded point, e.g. $X = \mathrm{Spec } k[x,y]/(xy, y^2)$ and $Z = \{0\}$ with reduced closed subscheme structure. –  Piotr Achinger Mar 10 '12 at 5:52
Also, it fails if you can find a vector bundle $E$ such that $H^1(X, E)\neq 0$ but $H^1(U, E|_U) = 0$ and $H^1(Z, E|_Z)=0$. This happens probably quite often, but I don't know if such examples exist in general. –  Piotr Achinger Mar 10 '12 at 6:06

Let me gather my comments into an answer. Take $X$ to be a smooth projective curve, $Z$ a closed point, $U$ its complement, $F=k$. We are asking if the map $\mathcal{O}_X\to \mathcal{O}_Z\oplus \mathcal{O}_U$ is a monomorphism in the derived category.

Take $E=\omega_X^{-1}$. Let $E[-1]\to \mathcal{O}_X$ be the map corresponding to the nontrivial element of $Ext^1(\omega_X^{-1}, \mathcal{O}_X) = H^1(X, \omega_X)=k$. Then since $H^1(U, E|_U)$ and $H^1(Z, E|_Z)$ are both zero (since $Z$ and $U$ are affine, the composition $E[-1]\to \mathcal{O}_X\to \mathcal{O}_Z\oplus\mathcal{O}_U$ is zero.

It may also happen that the map $\mathcal{O}\to \mathcal{O}_Z\oplus\mathcal{O}_U$ is not a monomorphism even in the category of sheaves, e.g. $X = \mathrm{Spec} k[x,y]/(xy, y^2)$ be the line with an embedded point, $Z$ be the reduced point 0 and $U$ the complement. Then the nilpotent global section $x$ of $\mathcal{O}_X$ maps to zero in both $\mathcal{O}_Z$ and $\mathcal{O}_U$.

-
What about $X=\mathbb P^1_k$, $Z$ a point, $U=\mathbb A^1_k$ the open complement, and $F=Q_\ell$ (constant sheaf) ?
Then the map $$Hom_{D(X)}(f^*F[-2],f^*F) \longrightarrow Hom_{D(X)}(f^*F[-2],i_*i^*f^*F) \oplus Hom_{D(X)}(f^*F[-2],j_*j^*f^*F)$$ gets identified with the restriction map $$H^2(\mathbb P^1_k, Q_\ell) \longrightarrow H^2(pt, Q_\ell) \oplus H^2(\mathbb A^1_k, Q_\ell)$$ which is clearly non injective.