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suppose $X$ is a smooth variety, with a finite open covering $U_i, i=1,....,k$. $F$ and $G$ are coherent sheaves on $X$, and $G$ is locally free.

Suppose on each $U_i$, there is a sheaf morphism $h_i: F_{U_i}-->G_{U_i}$.

Can this be extended on X ?

i started by taking push-forward, for each i: $$(h_i)_*: (h_i)_*F_{U_i} ---> (h_i)_*G_{U_i} = (h_i)_*O_{U_i} \otimes G. $$ Restrict the map $(h_i)_*$ on the subsheaf $F$ to get a morphism into right hand side.

Using the covering $U_i$ of $X$, can we conclude that the image of restriction, is in $\cap_i (h_i)_*\cal {O_{U_i}} \otimes G$, (which is equal to $G$) ?

share|improve this question
    
This is true if $F$ is locally free (and $G$ can be arbitrary, smoothness of $X$ is not important either) and if morphisms $h_i$ agree on intersections $U_i \cap U_j$. To prove replace $Hom(F,G)$ by $H^0(X,F^*\otimes G)$. –  Sasha Nov 21 '12 at 11:17
2  
I think the question -as it stands now- is not well phrased. For example you give no compatibility conditions on the $U_i\cap U_j$. What if $X=\mathbb{A}^1$, $k=2$, $U_1=U_2=\mathbb{A}^1$, $F=G=\mathcal{O}_X$, $h_1=\mathrm{id}_X$, $h_2=0$ ? –  Qfwfq Nov 21 '12 at 11:43
    
* Sorry, I meant: $h_1=1_{\mathcal{O}_X}$. –  Qfwfq Nov 21 '12 at 11:44

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