 Solution

Take $X=\mathbb{R}$, $U=(0,1)$, and let $\mathcal{F}$ be the skyscraper sheaf with value a nonzero abelian group $A$ at $p=3/4$. Cover $V=(-1,1)$ by $V_1=(-1,2/3)$ and $V_2=(1/2,1)$. The presheaf gives $(j^p\mathcal{F})(V)=0=(j^p\mathcal{F})(V_1)$ and $(j^p\mathcal{F})(V_2)=A$. A nonzero section on $V_2$ and the zero section on $V_1$ agree on the overlap, whose skyscraper sections vanish, but cannot be glued on $V$. Thus $j^p\mathcal{F}$ need not be a sheaf.

Sheafification preserves stalks. If $P\in U$, neighborhoods contained in $U$ are cofinal, so $(j^p\mathcal{F}{)}_P={\mathcal{F}}_P$. If $P\notin U$, no neighborhood of $P$ lies in $U$, so every term defining the presheaf stalk is zero. Therefore

Solved by gpt-5.6-sol high.