Let $k$ be an algebraically closed field of char $p\neq 0$, $W\_2(k)$ the witt vector of length 2. $C\_1$ a smooth projective curve over $W\_2(k)$, and $H\_1$ a vector bundle over $C\_1$. We denote $C\_0$ the smooth projective curve from mod $p$ reduction of $C\_1$ , and $H\_0$ the vector bundle from the reduction of $H\_1$ . Is the following statment true?


(1) If $H^0(C\_0,H\_0)\neq 0$, then $H^0(C\_1,H\_1)\neq0$. 


If (1) is ture, I may go on asking:


(2) The map $H^0(C\_1,H\_1)\to H^0(C\_0,H\_0)$ is surjective. 


Thank you!



global section of vector bundle and reduction
Here is a counterexample to (2).


Let $H\_1$ be the lifting of the trivial line bundle on $C\_0$ and suppose that 
$H\_1\not\simeq{\cal O}\_{C\_1}$. Examples of such line bundles $H\_1$ may be produced using 
the Picard scheme of $C\_1$ over $W\_2(k)$. I contend that the morphism 
$H^0(C\_1,H\_1)\to H^0(C\_0,H\_0)$ vanishes. To see this, let $\sigma\_1\in H^0(C\_1,H\_1)$. 
This corresponds to a morphism of sheaves $\sigma\_1:{\cal O}\_{C\_1}\to H\_1$. Let 
$K\_1$ be the kernel of $\sigma\_1$ and ${\rm CK}\_1$ be the cokernel of $\sigma\_1$. Let 
let $K\_0$ (resp. ${\rm CK}\_0$) be the reduction mod. $p$ of $K\_1$ (resp. ${\rm CK}\_1$). 
The reduction 
mod. $p$ of $\sigma\_1$ gives a morphism $\sigma\_0:{\cal O}\_{C\_0}\to H\_0\simeq{\cal O}\_{C\_0}$. 
We want to show that $\sigma\_0=0$. To get a contradiction, suppose that $\sigma\_0\not=0$. 
Then $\sigma\_0$ is an isomorphism, since $C\_0$ is proper over $k$ and the source and target 
of $\sigma\_0$ are trivial. 
Since the tensor product is right-exact, we deduce that ${\rm CK}\_0$ vanishes; 
but this implies that ${\rm CK}\_1$ vanishes. Now using the fact that $H\_1$ is locally free, we deduce likewise that $K\_0$ vanishes and hence that $K\_1$ vanishes. This shows that $\sigma\_1$ is an isomorphism, which 
contradicts the assumption on $H\_1$. Hence $\sigma\_0=0$, which is what we wanted. 


