Compute $H^1(X,\Bbb{Z}_U)$ Let $X = \mathbb{A}^1_k$ with $k$ infinite and $U = X - \{P,Q\}$ and $\mathbb{Z}_U= i_{!}(\mathbb{Z}|_U)$, $\Bbb{Z}$ the constant sheaf. I want to say that $H^1(X,\mathbb{Z}_U) \neq 0$. If it is zero we see exact sequence
$$0 \to H^0(X,\mathbb{Z}_U) \to H^0(X,\mathbb{Z}) \to H^0(X,\mathbb{Z}_{\{P,Q\}} ) \to0$$
where $\mathbb{Z}_Y$ is $j_\ast(\mathbb{Z}|_{\{P,Q\}})$. The middle term is $\mathbb{Z}$ but can I say right term is $\mathbb{Z} \oplus \mathbb{Z}$? It seems like it's actually $\mathbb{Z}$. I am confused.
 A: Since your sheaf $\mathbb{Z}_{\{P,Q\}}$ is equal to $j_\ast(\mathbb{Z}|_{\{P,Q\}})$, we have : $$H^0(X,\mathbb{Z}_{\{P,Q\}} ) =H^0(X, j_\ast(\mathbb{Z}|_{\{P,Q\}}))\stackrel {definition}{=}H^0(\{P,Q\}, \mathbb{Z}_X|_{\{P,Q\}})\stackrel {!}{=} H^0(\{P,Q\}, \mathbb{Z})=\mathbb{Z}\oplus \mathbb{Z}$$ This indeed shows that your displayed sequence is not exact (since $H^0(X,\mathbb{Z})=\mathbb  Z$ and  $\mathbb Z$ cannot be surjected onto $\mathbb{Z}\oplus \mathbb{Z}$) and thus that necessarily  $H^1(X,\mathbb{Z}_U) \neq 0$.     
Edit
It is actually fairly easy to see that  $ H^1(X,\mathbb{Z})=0 $ (because constant sheaves are flasque and thus acyclic on irreducible spaces), and that $H^1(X,\mathbb{Z}_U) \cong \mathbb Z$ since it is the cokernel of the restriction morphism  $ H^0(X,\mathbb{Z})= \mathbb{Z} \to H^0(\{P,Q\},\mathbb{Z} )=\mathbb{Z} \oplus \mathbb{Z}: n\mapsto (n,n)$  
New edit
As discussed in the comments I have used that the restriction of a constant sheaf $A_X$ on a space $X$ to any subspace $Y$ is the constant sheaf $A_Y$  to deduce $H^0(\{P,Q\}, \mathbb{Z_X}|_{\{P,Q\}})\stackrel {!}{=} H^0(\{P,Q\}, \mathbb{Z})$
