suppose we have an exact sequence of locally free sheaves on a smooth variety: $$ 0\rightarrow E \rightarrow F \stackrel{f}{\rightarrow} G \rightarrow 0. $$ On an open subset $U\subset X$, this splits. Suppose $H$ is a proper subsheaf of $G$. Then on $U$, regard $H$ as a subsheaf of $F$. take its coherent extension $H'\subset f^{-1}(H) $ on $X$. It is torsion free. Then the kernel of $H'\rightarrow H$ on $X$ is subsheaf of $E$ and supported on $X-U$. hence the kernel is zero since $E$ is locally free.

Assume that $H$ is reflexive. Is it correct that the cokernel of $H'\rightarrow H$ is supported in codimension two subset (it is a torsion sheaf). ?

(take double dual and use that $H'\subset H'^{**}$ is isomorphism outside codim two subset)


1 Answer 1


No that is not true. Already there are counterexamples on $X=\mathbb{P}^1$. Consider the standard short exact sequence, $$ 0 \to \mathcal{O}(-1) \to \mathcal{O}\oplus \mathcal{O} \to \mathcal{O}(+1) \to 0,$$ and take $H=G=\mathcal{O}(+1)$. Every torsion-free, coherent subsheaf $H'$ of $\mathcal{O}\oplus \mathcal{O}$ is automatically locally free. So your sheaf $H'$ is an invertible sheaf that admits an injective sheaf homomorphism to $\mathcal{O}\oplus \mathcal{O}$. By taking the transpose map, $(H')^\vee$ has nonzero global sections. Thus $(H')^\vee$ has nonnegative degree, i.e., $H'$ has nonpositive degree. Thus the induced morphism $H'\to H$ fails to be an isomorphism already in codimension $1$.


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