Let $f : X \to \operatorname{Spec}(R)$ be a flat, projective morphism with reduced fibers, where $R =\mathbb{C}[[t]]$. One may assume that the dualizing sheaf of the special fiber is trivial. Let $\mathcal{F}$ be a coherent sheaf on $X$, not necessarily flat over $R$. Let $\mathcal{G}$ be the sheaf of $t$-torsion sections of $\mathcal{F}$, i.e $\mathcal{G} = \operatorname{ker}(\mathcal{F} \xrightarrow{t} \mathcal{F})$.

Does $H^1(X,\mathcal{F}) = 0$ imply that $H^1(X,\mathcal{G})= 0$?


$\def\C{\mathbb{C}}\def\OO{\mathcal{O}}\def\cL{\mathcal{E}}\def\cF{\mathcal{F}}$Let $X_0$ be a variety over $\C$ and $X=X_0\times_{\C}R$. Denote by $i:X_0\to X$ the obvious closed immersion.

Let $\cL$ be a coherent sheaf on $X_0$. We will construct an extension $0\to i_*\cL\to \cF\xrightarrow{p} i_*\OO\to 0$. Such extensions are classified by $$Hom_{D(\OO_X)}(i_*\OO_{X_0},i_*\cL[1])=Hom_{D(\OO_{X_0})}(Li^*i_*\OO_{X_0},\cL[1])=\\ Hom_{D(\OO_{X_0})}(Li^*(cone(\OO_{X}\xrightarrow{t}\OO_{X})),\cL[1])=Hom_{D(\OO_{X_0})}(\OO_{X_0}[1]\oplus \OO_{X_0},\cL[1])$$ Given such extension, there is a long exact sequence $$0\to H^0(X_0,\cL)\to H^0(X,\cF)\to H^0(X_0,\OO_{X_0})\xrightarrow{a} H^1(X_0,\cL)\to H^1(X,\cF)\to H^1(X_0,\OO_{X_0})$$ The map $a$ carries the section $1\in H^0(X_0,\OO_{X_0})$ to the class in $H^1(X_0,\cL)=Hom_{D(X_0)}(\OO_{X_0},\cL[1])$ which is the projection of the class in $Hom_{D(\OO_{X_0})}(\OO_{X_0}[1]\oplus \OO_{X_0},\cL[1])$ corresponding to our extension.

Let's now analyze $\cF$ itself. Since it is an extension of sheaves killed by $t$, it is killed by $t^2$. Multiplication by $t$ on $\cF$ kills $i_*\cL$ and defines a map of sheaves $f:i_*\OO_{X_0}\to i_*\cL$ which is, when considered as an element of $Hom_{D(\OO_{X_0})}(\OO_{X_0}[1],\cL[1])$ is again another projection of the class corresponding to the exntension. Thus, $\ker(\cF\xrightarrow{t}\cF)=i_*\cL+p^{-1}(\ker(\OO_{X_0}\to \cL))$. Note that $\cF$ is usually not going to be flat over $\C[t]/t^2$ since $\mathrm{im}(\cF\xrightarrow{t}\cF)=\mathrm{im}(i_*\OO_{X_0}\to i_*\cL)$ and for a flat sheaf $\ker t$ and $\mathrm{im}\, t $ should be equal(they are going to be equall iff $f$ is an isomorphism).

So, let's pick an irreducible $X_0$ and $\cL$ such that $H^0(X_0,\cL)\neq 0, H^1(X_0,\cL)=\C, H^1(X_0,\OO_{X_0})=0$, e.g. $X_0=\mathbb{P}^1,\cL=\OO\oplus\OO(-2)$(probably, one can also find such situattion on a variety with trivial dualizing sheaf, say, on a K3 surface). Using a non-zero section $\OO_{X_0}\to \cL$ and a non-zero cohomology class $\OO_{X_0}\to \cL[1]$, construct a class in $Hom_{D(\OO_{X_0})}(\OO_{X_0}[1]\oplus \OO_{X_0},\cL[1])$ and take the corresponding extension $\cF$. Since the map $f:i_{*}\OO_{X_0}\to i_*\cL$ is injective(it corresponds to a non-zero section), $\mathcal{G}=i_*\cL$. Since we took a non-zero cohomology class, which, moreover, generates $H^1(X_0,\cL)$, the connecting homomorphism $a$ is surjective so $H^1(X,\cF)=0$, but $H^1(X,\mathcal{G})=H^1(X_0,i_*\cL)\neq 0$.


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