Cohomology of sheaf extended by zero Let $X$ be a projective scheme of pure dimension $1$. Let $U$ be a open subscheme and $j:U \to X$ the open immersion. Let $\mathcal{F}$ be a coherent sheaf on $U$.
Denote by $j_!(\mathcal{F})$ the extension of $\mathcal{F}$ by zero (as mentioned in Hartshorne, Algebraic geometry Ex. II.$1.19$). Is it true that $H^1(U,\mathcal{F})=H^1(X,j_!(\mathcal{F}))$?
 A: No, this is not true. This works for extensions from closed subsets. Take for $X$ an irrational curve, so that $H^1(X;\mathcal{O}_X)\ne0$, and let $U=X\setminus\text{point}$ and $\mathcal{F}=\mathcal{O}_U$, which is the restriction of $\mathcal{O}_X$. You immediately get $H^1(X,j_!\mathcal{F})\twoheadrightarrow H^1(X;\mathcal{O}_X)$, whereas $H^1(U,\mathcal{F})=0$.
A: Actually, this is almost never true. 
Claim Let $X$ be a separated one-dimensional scheme of finite type over an algebraically closed field. Let $U\subseteq X$ be a dense open set that does not contain any irreducible component of $X$. Further let $\mathscr F$ be a coherent sheaf on $X$. Then 


*

*$H^1(U,\mathscr F|_U)=0$, and 

*$H^1(X,j_!(\mathscr F|_U))\twoheadrightarrow H^1(X,\mathscr F)$ is surjective.


In particular, if $H^1(X,\mathscr F)\neq 0$, then $H^1(U,\mathscr F|_U)\neq H^1(X,j_!(\mathscr F|_U))$
Proof The assumptions imply that 


*

*$U$ is affine, and 

*$Z=X\setminus U$ is zero-dimensional.


The above 1. implies 1. in the Claim and via the short exact sequence
$$
0\to j_!(\mathscr F|_U) \to \mathscr F \to j_*(\mathscr F|_Z) \to 0
$$
the above 2. implies 2. in the Claim. $\square$
