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$\def\sO{\mathcal{O}} \def\sI{\mathcal{I}}$Given a locally ringed space $Y$ and an ideal sheaf $\sI\subset\sO_Y$, we can consider the closed subspace of $Y$ cut off by $\sI$, i.e., the closed immersion of locally ringed spaces $Z\to Y$ where $Z=\operatorname{Supp}(\sO_Y/\sI)$ and $\sO_Z=\sO_Y/\sI|_Z$. Now let $f:X\to Y$ be a locally closed immersion of locally ringed spaces. By definition, this means that $f^{-1}\sO_Y\to\sO_X$ is surjective and that on spaces $f$ is a homeomorphism of $X$ onto a locally closed subset of $Y$. What is the closed subspace $Z$ of $Y$ cut off by the ideal sheaf $\sI=\ker(\sO_Y\to f_*\sO_X)$?

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$\def\sO{\mathcal{O}}$We claim $Z=\overline{f(X)}$. On the one hand, it is clear that $f(X)\subset Z$ (there is open $U\subset Y$ containing $f(X)$ as a closed subset; thus $Z\cap U=\operatorname{Supp}(\mathcal{O}_U/\mathcal{I}|_U)=f(X)$), whence $\overline{f(X)}\subset Z$. Conversely, let $z\in Z$ and suppose $V\subset Y$ is an open neighborhood of $z$. Then $(\mathcal{I}|_V)_z=\mathcal{I}_z\neq \sO_{X,z}=\sO_{V,z}$. In particular, there is an open neighborhood $W\subset V$ of $z$ with $\mathcal{I}(W)\neq\sO_Y(W)$; thus, $\mathcal{O}_X(f^{-1}(W))\neq 0$, so $\varnothing\neq W\cap f(X)\subset V\cap f(X)$. This means $z\in\overline{f(X)}$.

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