Extension by zero is left adjoint to restriction and is exact on abelian sheaves

Let $\mathrm{\Phi }:j_{!}\mathcal{F}\to \mathcal{G}$ be a morphism on $X$. If $W\subseteq U$ is open, then [F2] gives $(j_{!}\mathcal{F})(W)=\mathcal{F}(W)$, so the components ${\mathrm{\Phi }}_W$ define a morphism $\mathcal{F}\to j^{-1}\mathcal{G}$ on $U$.

Conversely, let $\mathrm{\Psi }:\mathcal{F}\to j^{-1}\mathcal{G}$ be a morphism on $U$. For an open set $V\subseteq X$ and a section $s\in (j_{!}\mathcal{F})(V)$, use $\mathrm{\Psi }$ to obtain a section on $V\cap U$. For each point $x\in V\setminus U$, the support condition in [F2] gives an open neighbourhood $W_x\subseteq V$ with $s{\mathrm{\mid }}_{W_x\cap U}=0$, so we may assign the zero section of $\mathcal{G}$ on $W_x$. These local sections agree on overlaps, so they glue uniquely to a section of $\mathcal{G}(V)$. This defines a morphism $j_{!}\mathcal{F}\to \mathcal{G}$.

Let $x\in X$. If $x\in U$, then the neighbourhoods of $x$ that lie inside $U$ are cofinal among all neighbourhoods of $x$, so $(j_{!}\mathcal{F}{)}_x\cong {\mathcal{F}}_x$. If $x\mathrm{\notin }U$ and $s\in (j_{!}\mathcal{F})(V)$ for some neighbourhood $V$ of $x$, then [F2] gives an open neighbourhood $W_x\subseteq V$ with $s{\mathrm{\mid }}_{W_x\cap U}=0$, so the germ of $s$ at $x$ is zero. Therefore $(j_{!}\mathcal{F}{)}_x=0$.

The constructions in steps 1.1 and 1.2 are inverse, because on opens contained in $U$ they both recover the original morphism on $\mathcal{F}$, and outside $U$ the support condition forces the extension to vanish locally. Therefore $j_{!}$ is left adjoint to $j^{-1}$.

Apply step 1.3 to a short exact sequence on $U$. At points of $U$ the stalk sequence after $j_{!}$ is the original exact stalk sequence, and outside $U$ it is $0\to 0\to 0\to 0$. Hence [L1] implies that $j_{!}$ preserves exact sequences.