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Let $f: X \to Y$ be a morphism of finite type of schemes over a field $k$. Assume that there exists $x \in X$ such that $\mathcal{O}_{Y, f(x)} \to f_* \mathcal{O}_{X, x}$ is an isomorphism. Is it true that there exist open subschemes $f(x) \in U \subset Y$ and $x \in V \subset X$ such that $f{|_V}: V \to U$ is an isomorphism? Could we drop the finiteness condition?

Probably, a natural generalization is the following: assume that $(R_i)_{i \in I}$ and $(S_j)_{j \in J}$ are directed systems. Suppose that we have maps $R_i \to colim(S_j)$ and for every $i \in I$ there exists $j \in J$ and the map $R_i \to S_j$ compatible with the map $q: colim(R_i) \to colim(S_j)$. Assume the $q$ is an isomorphism. Is it true that there exists a pair of indices $i, j$ such that $R_i \to S_j$ is an isomorphism? If no, what conditions one should impose to make it true?

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    For your generalization, I think it’s no. Consider, for positive $i$, $R_i=k[x_0,\ldots,x_{2i}]$ and $S_i=k[x_0,\ldots,x_{2i+1}]$. For the original question, you would need to clarify what you mean by $f_*\mathcal{O}{X,x}$ (where are the parentheses?!). If that’s just $f^{\sharp}: \mathcal{O}{Y,f(x)} \rightarrow \mathcal{O}_{X,x}$ being an isomorphism, then the statement is true. – Aphelli Feb 15 '22 at 22:19
  • Your hypothesis looks strange. Do you mean to assume $O_{Y, f (x)} \to O_{X, x}$, or equivalently, $(f^* O_Y)x \to O{X, x}$ is an isomorphism? – Zhen Lin Feb 15 '22 at 22:23
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    Your first question is essentially Hartshorne I.4.7, which has been dealt with on MSE before. The idea is that we can lift the isomorphism to an isomorphism on an affine open neighborhood by inverting at most one element in each of $k[V]$ and $k[U]$. The finiteness condition cannot be dropped: consider Spec of $k[t]\to k[t]_{(t)}$. Any open subscheme of the source is finite type over $k$, while no open subscheme of the target is. – KReiser Feb 15 '22 at 22:29

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