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Can someone me say if this (perhaps obvious!) claim is true:

let $f:X\rightarrow S$ be an open, surjective morphism of complex spaces reduced or without embedded components and with $n$-pure dimensional fibers. Let $F$, $G$ be a coherent sheaves with $S$-depth bigger than $2$ and s.t:

1) $G$ is $S$-flat

2) there is a canonical injective morphism $F\rightarrow G$

3) $F$ and $G$ are locally isomorphics.

Question: Is $F$ $S$-flat too?

Rk: Of course, the answer is yes if the Coker of 2) is $S$-flat.

Thanks.

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    $\begingroup$ $S$-flatness is a local property, so 1) and 3) imply that $F$ is $S$-flat, without any further assumptions. Or am I missing something? $\endgroup$ Nov 22, 2010 at 8:55
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    $\begingroup$ @Harry: Locally isomorphic does not mean that there is a global morphism that gives the local isomorphism. For instance, any two line bundles are locally isomorphic, but there may not be a morphism between them. Or there may be one, say one could be a subsheaf of the other, but the local isomorphism is not given by the global morphism. $\endgroup$ Nov 22, 2010 at 9:31
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    $\begingroup$ Actually I cannot think any example where one requires locally isomorphic objects to have a global morphism. For instance, differentiable manifolds are locally diffeomorphic to $\mathbb{R}^n$, regardless of their global morphisms to or from $\mathbb{R}^n$. $\endgroup$ Nov 22, 2010 at 10:22
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    $\begingroup$ Then as you said, flatness is local on $X$ and so is $S$-flatness. I still don't see the problem. $\endgroup$ Nov 22, 2010 at 15:39
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    $\begingroup$ @Harry: I don't know where this is a "standard" definition. How about locally free? Do you need a morphism to or from a free sheaf for that? $\endgroup$ Nov 23, 2010 at 8:54

1 Answer 1

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For a flat map $f$, we can see the $S$-flatness of $F$ as follow:

Let $x\in X$ and denote by: $A:=O_{S,f(x)}$, $B:=O_{X,x}$, ${\rm F}:=F_{x}$ and ${\rm G}:=G_{x}$. The $S$-flatness of the coherent sheaf $F$ in $x$ is, by definition, the $A$-flatness of the module (of finite type) ${\rm F}$ or, equivalently, the left exactness of the functor $K\rightarrow {\rm F}\otimes_{A} K$ where $K$ is $A$-module of finite type. Take a left exact sequenz of $A$-modules $$0\rightarrow K\rightarrow L$$ which give us a morphism ${\rm F}\otimes_{A}K\rightarrow {\rm F}\otimes _{A} L$ or equivalently a morphism

$${\rm F}{\otimes_{B}}(B\otimes_{A} K)\rightarrow {\rm F}\otimes_{B}(B\otimes_{A} L)$$

We can replace ${\rm F}$ by ${\rm G}$ at this moment and see that the injectivity of the desired map is obtained if we assume $B$ is $A$-flat module because $B\otimes_{A} K\rightarrow B\otimes_{A} L$ is injective and ${\rm G}$ is $A$-flat.

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