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 $H^1(U,\mathcal{F})=H^1(X,j\_!(\mathcal{F}))$?



Cohomology of sheaf extended by zero
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 


1. $H^1(U,\mathscr F|\_U)=0$, and
2. $H^1(X,j\_!(\mathscr F|\_U))\twoheadrightarrow H^1(X,\mathscr F)$ 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 


1. $U$ is affine, and
2. $Z=X\setminus U$ 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. $\square$


