Let $X$ be a ringed space, and $J$ be a sheaf of ideals of the structure sheaf. Define, $Y = \{x\in X ~ | ~ J_x \not = \mathcal{O}_x\}$, this is a closed set. We have an inclusion $i:Y\to X$. Is there categorical justification that, $i_*i^*(\mathcal{O}/J) = \mathcal{O}/J $?
I really do not want to work with the tedious constructions to justify that.
I know that one has the following exact sequence, $$ 0 \to i_!i^*\mathcal{O}/J \to \mathcal{O}/J \to i_*i^*\mathcal{O}/J \to 0 $$
Now check that $F = i_!i^*\mathcal{O}/J$ has zero stalk for every $x\in X$. If $x\not \in Y$, then $F_x = 0$. If $x\in Y$, then $F_x = \mathcal{O}_x/J_x \not = 0$. And this is unforunate as it prevents the natural map $\mathcal{O}/J \to i_*i^*\mathcal{O}/J$ from being an isomorphism.
Your exact sequence is wrong. One has the exact sequence (See Exc. I.1.19 in Hartshorne)
$$0 \to j_{!}j^* \mathcal O/J \to \mathcal O/J \to i_*i^*\mathcal O/J \to 0,$$
where $i: Y \to X$ is the inclusion of the closed subset and $j: X \setminus Y =: U \to X$ is the inclusion of the open complement.
For $x \in Y$, we have $(j_{!}j^* \mathcal O/J)_x=0$.
For $x \in U$, we have $(j_{!}j^* \mathcal O/J)_x=(j^* \mathcal O/J)_x = (\mathcal O/J)_{j(x)} = (\mathcal O/J)_x=\mathcal O_x/J_x=0$, since $U$ is precisely the locus, where this quotient vanishes.
Hence, we obtain that $\mathcal O/J \to i_*i^*\mathcal O/J$ is an isomorphism.