I have a question ;
Let $ U \overset{i} \hookrightarrow Y \overset{j} \hookrightarrow Z $ , where i is closed open inclusion ( ; i.e, closed open subscheme), and j is closed inclusion, and $ k:= j \circ i : U \hookrightarrow Z $ is closed immersion. Then now define $ \mathcal{J} := ker(\mathcal{O}_{Z} \overset {j^{b}} \rightarrow j_{*} \mathcal{O}_{Y} ) $ and $ \mathcal{I} := ker(\mathcal{O}_{Z} \overset {k^{b}} \rightarrow k_{*} \mathcal{O}_{Y}|_{U} ) $ ; i.e. the sheaf of ideals defining j,k respectively.
($j^{b}, k^{b} $ are surjective since j, k are closed immersion)
Q. Then $\mathcal{J} = \mathcal{I}$ ?
This question originates from next proof
I can't understand the above underlined statement. If my question is true, then it seems to be possible to deduce the underlined statement, since then $(f \times f)^{*} \ \mathcal{I}$ becomes also sheaf of ideal defining $\Delta _{X/S} = q\Delta_{X/Y}$ .
Is it really true?
If $U$ is the spectrum of $\mathbb{C}=\mathbb{C}[x]/(x)$, $Y$ that of $\mathbb{C}[x]/(x(x-1))$, $Z$ that of $\mathbb{C}[x]$, then (unless I’m mistaken) $\mathcal{I}$ (resp. $\mathcal{J}$) is the quasi-coherent sheaf of ideals on $Z$ whose global sections are the ideal $(x)$ (resp. $(x(x-1))$) – so they’re not equal.
If you watch the proof closely, what they’re using is that (in your setting) $k^*\mathcal{I}=k^*\mathcal{J}$.
This one is true, because obviously $\mathcal{J}$ is a subsheaf of $\mathcal{I}$ and if $x \in U$, $\mathcal{I}_x=\mathcal{J}_x$, morally because $U$ is an open subset of $Y$, and (more rigorously) because it holds when $Z$ is affine.