I am trying to solve the exercise 2.4 chapter III in Hartshorne's "Algebraic Geometry". For this I would like to prove for a sheaf $F$ of abelian groups on a topological space $X$ and $U$ open subset of $X$ with complement $Z$ the following $$ H^{i}_{Z}(X,j_{.}(F\mid_{U})) = 0 \text{ for all }i\geq 0$$
where $j:U \longrightarrow X$ the inclusion and $j_{.}$ meaning the extension of a sheaf on $U$ to a sheaf on $X$ by zero. I have proven exercise 2.3 chapter III, so i can use this.
Does anyone know a nice proof for this? Or maybe a counterexample if this doens't hold? I was able to reduce/adapt this case to proving $H^{i}(X,j_{.}(F\mid_{U})) \cong H^{i}(U,F\mid_{U})$. Also for it is quite clear why $H^{0}_{Z}(X,j_{.}(F\mid_{U})) =0$, it is just for the higher order groups I dont have an argument.
Thanks in advance!
This is false, and in fact it almost never holds...
An easy counter-example is given by $X=\mathbb{R}^2$, $Z=\{(0,0)\}$ and $U=\mathbb{R}^2\setminus\{(0,0)\}$ with $\mathcal{F}=\mathbb{Z}$ the constant sheaf. I claim that $H^1_Z(X,j_!\mathbb{Z})\neq 0$.
To see this, we will use two long exact sequences : for any $\mathcal{F}\in\operatorname{Sh}(X)$ there is a long exact sequence : $$...H^i_Z(X,\mathcal{F})\to H^i(X,\mathcal{F})\to H^i(U,\mathcal{F}|_U)\to H^{i+1}_Z(X,\mathcal{F})\to ...$$ which is functorial in $\mathcal{F}$, and in particular functorial with the long exact sequence associated to $$ 0\to j_!\mathbb{Z}\to \mathbb{Z}\to i_*\mathbb{Z}\to 0$$
This gives the diagram : $$ \require{AMScd} \begin{CD} 0@>>>0@>>>H^0_Z(X,i_*\mathbb{Z})@>>>H^0(X,i_*\mathbb{Z})@>>>H^0(U,i_*\mathbb{Z})\\ @VVV@VVV@VVV@VVV@VVV\\ H^0(X,j_!\mathbb{Z})@>>>H^0(U,\mathbb{Z})@>>>H^1_Z(X,j_!\mathbb{Z})@>>>H^1(X,j_!\mathbb{Z})@>>> H^1(U,\mathbb{Z})\\ @VVV@|@VVV@VVV@|\\ H^0(X,\mathbb{Z})@>>>H^0(U,\mathbb{Z})@>>>H^1_Z(X,\mathbb{Z})@>>> H^1(X,\mathbb{Z})@>>>H^1(U,\mathbb{Z})\\ @VVV@VVV@VVV@VVV@VVV\\ H^0(X,i_*\mathbb{Z})@>>>H^0(U,i_*\mathbb{Z})@>>>H^1_Z(X,i_*\mathbb{Z})@>>> H^1(X,i_*\mathbb{Z})@>>>H^1(U,i_*\mathbb{Z}) \end{CD} $$ Now using some easy computations and standard results of cohomology with coefficients in the constant sheaf $\mathbb{Z}$. $$ \require{AMScd} \begin{CD} 0@>>>0@>>>\mathbb{Z}@>>>\mathbb{Z}@>>>0\\ @VVV@VVV@VVV@VVV@VVV\\ H^0(X,j_!\mathbb{Z})@>>>\mathbb{Z}@>>>H^1_Z(X,j_!\mathbb{Z})@>>>H^1(X,j_!\mathbb{Z})@>>>\mathbb{Z}\\ @VVV@|@VVV@VVV@|\\ \mathbb{Z}@>>>\mathbb{Z}@>>>0@>>> 0@>>>\mathbb{Z}\\ @|@VVV@VVV@VVV@VVV\\ \mathbb{Z}@>>>0@>>>H^1_Z(X,i_*\mathbb{Z})@>>> 0@>>>0 \end{CD} $$ This shows that $\mathbb{Z}$ injects into $H^1_Z(X,j_!\mathbb{Z})$, thus this group in not zero. (If we push the diagram a bit further, we can show that this is in fact an isomorphism).