global section of vector bundle and reduction 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!
 A: 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. 
