Extension by zero and the closed complement: a short exact sequence

Let $\mathcal{G}:=j_{!}(\mathcal{F}{\mathrm{\mid }}_U)$. For $x\in U$ the open neighbourhoods $V$ of $x$ with $V\subseteq U$ are cofinal among all open neighbourhoods of $x$, and on them $(j_{!}(\mathcal{F}{\mathrm{\mid }}_U))(V)=(\mathcal{F}{\mathrm{\mid }}_U)(V)=\mathcal{F}(V)$ by [F1] and [F2]; hence ${\mathcal{G}}_x={\mathrm{colim}}_{x\in V\subseteq U}\mathcal{F}(V)={\mathcal{F}}_x$ by [F3], the colimit over the cofinal subdiagram agreeing with the stalk of $\mathcal{F}$.

Let $x\in X\setminus U$ and let $s\in \mathcal{G}(V)$ be a section over an open neighbourhood $V$ of $x$. Since $x\in V\setminus U$, the local description of [F1] provides an open $W_x\subseteq V$ with $x\in W_x$ and $s{\mathrm{\mid }}_{W_x\cap U}=0$, and the restriction of $s$ to $W_x$ is the zero element of $\mathcal{G}(W_x)$; by [F4] applied to the filtered diagram of section groups over the neighbourhoods of $x$ [F3], the image of $s$ in ${\mathcal{G}}_x$ is zero. Every element of ${\mathcal{G}}_x$ has such a representative, so ${\mathcal{G}}_x=0$ for every $x\mathrm{\notin }U$.

Let $\mathcal{H}:=i_{\ast }(\mathcal{F}{\mathrm{\mid }}_Z)$, a sheaf of abelian groups on $X$ by [F5]. For $z\in Z$ the traces $V\cap Z$ of the open neighbourhoods $V$ of $z$ in $X$ are cofinally the open neighbourhoods of $z$ in $Z$ [F7], so by [F5] and [F3] the stalk is ${\mathcal{H}}_z\cong (\mathcal{F}{\mathrm{\mid }}_Z{)}_z\cong {\mathcal{F}}_z$ by [F9]. For $x\mathrm{\notin }Z$ the open set $X\setminus Z$ contains $x$, and $(V\cap (X\setminus Z))\cap Z=\mathrm{\varnothing }$ for every open $V\ni x$, so the colimit of [F3] is computed by the subdiagram of neighbourhoods meeting $Z$ in the empty set, where $\mathcal{H}$ has the constant value $(\mathcal{F}{\mathrm{\mid }}_Z)(\mathrm{\varnothing })$, a singleton by [F6] and hence the zero group; thus ${\mathcal{H}}_x=0$ for every $x\mathrm{\notin }Z$.

I record how the two maps of the displayed sequence act on stalks. The map $j_{!}(\mathcal{F}{\mathrm{\mid }}_U)\to \mathcal{F}$ is the adjoint transpose of the identity of $\mathcal{F}{\mathrm{\mid }}_U$; on a section over $V\subseteq U$ it is the identity of $\mathcal{F}(V)$ under the identifications of [F1] and [F2], so for $x\in U$ its stalk is the identification ${\mathcal{G}}_x={\mathcal{F}}_x$ of [step 1.1]. The map $\mathcal{F}\to i_{\ast }(\mathcal{F}{\mathrm{\mid }}_Z)$ is the unit of the adjunction [F8]; its stalk at $z\in Z$ is the canonical identification ${\mathcal{F}}_z\cong (\mathcal{F}{\mathrm{\mid }}_Z{)}_z\cong {\mathcal{H}}_z$ of [step 1.3] and [F9], because the unit is adjunct to the identity of $i^{-1}\mathcal{F}$, and for $x\mathrm{\notin }Z$ the target stalk ${\mathcal{H}}_x$ vanishes by [step 1.3], so that stalk map is ${\mathcal{F}}_x\to 0$.

At a point $x\in U$ the stalk sequence of the displayed sequence is $0\to {\mathcal{G}}_x={\mathcal{F}}_x\to {\mathcal{F}}_x\to {\mathcal{H}}_x=0\to 0$, the first map being the identity by [step 2.1] and the last the zero map, so it is exact; at a point $z\in Z$ it is $0\to {\mathcal{G}}_z=0\to {\mathcal{F}}_z\to {\mathcal{H}}_z\cong {\mathcal{F}}_z\to 0$, the last map being an isomorphism by [step 2.1], so it is exact as well. By [F10] the displayed sequence of sheaves is exact at $j_{!}(\mathcal{F}{\mathrm{\mid }}_U)$, at $\mathcal{F}$ and at $i_{\ast }(\mathcal{F}{\mathrm{\mid }}_Z)$, that is, it is a short exact sequence; this proves clause 1.

Let $\mathcal{F}={\mathbb{Z}}_X$ be the constant sheaf of [F11]. For $W\subseteq U$ open, the definition of the restriction gives $({\mathbb{Z}}_X{\mathrm{\mid }}_U)(W)={\mathbb{Z}}_X(W)$ [F2], and ${\mathbb{Z}}_X(W)$ is identified with the group of locally constant $\mathbb{Z}$-valued functions $W\to \mathbb{Z}$ by [F11]; the same group is ${\mathbb{Z}}_U(W)$, and the identifications are those of the canonical isomorphisms of [F11] for the spaces $X$ and $U$, so they are compatible with restrictions in $W$. Hence ${\mathbb{Z}}_X{\mathrm{\mid }}_U$ is canonically isomorphic to ${\mathbb{Z}}_U$, and the sequence of clause 1 takes the form $0\to j_{!}({\mathbb{Z}}_U)\to {\mathbb{Z}}_X\to i_{\ast }(\mathcal{F}{\mathrm{\mid }}_Z)\to 0$ up to this isomorphism, which proves clause 2. ∎