Do we have exact sequence $0\to i_*F\to F\to j_!F\to 0$

Question

Suppose $X$ is a topological space, $F$ is a sheaf of abelian groups, $i$ be inclusion of a closed subset $Z$, $j$ be the inclusion of its complement. We do have exact sequence $0\to j_!F|_U\to F\to i_*F|_Z\to 0$. Do we also have exact sequence $0\to i_*F|_Z\to F\to j_!F|_U\to 0$? ($j_!$ is the compact supported pushforward, or the sheaf defined by $j_!F|_U(V)=0$ if $V\cap Z\neq\emptyset$, $j_!F|_U(V)=F(U)$ if $U\subseteq X-Z$)

Answer 1

In general there is no morphism $F\rightarrow j_! F$ which induce an isomorphism on the stalks at the points $x\in U$. For example, take $F=\mathbb{Z}$ the constant sheaf on $X=\mathbb{C}$. Take $U=\mathbb{C}^*$. You have $\Gamma(\mathbb{C},j_!\mathbb{Z})=0$ since the map $\mathbb{Z}\rightarrow i_*\mathbb{Z}$ induce an isomorphism on global sections.

Now, if $x\in U$, the following diagram is commutative : $$ \require{AMScd} \begin{CD} \Gamma(\mathbb{C},\mathbb{Z}) @>>> \Gamma(\mathbb{C},j_!\mathbb{Z})\\ @VVV @VVV \\ \mathbb{Z}_x @>>> (j_!\mathbb{Z})_x \end{CD} $$ Because $\Gamma(\mathbb{C},j_!\mathbb{Z})=0$, the map $\Gamma(\mathbb{C},\mathbb{Z})\rightarrow (j_!\mathbb{Z})_x$ is zero. But the left vertical map is an isomorphism, so the bottom map is also zero, so not an iso as wanted. In fact, in this particular case, the only morphism from $\mathbb{Z}$ to $j_!\mathbb{Z}$ is zero.

The exact sequence $$ 0\rightarrow j_! F_{|U}\rightarrow F\rightarrow i_* F_{|Z}\rightarrow 0$$ almost never splits. It splits if $U$ is also closed, or if $F$ is of the form $j_!F \oplus i_*G$, but these are very particular cases.