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Note: I originally asked this on MSE without any success.

Let $f\colon X\to Y$ be a morphism of schemes for which $f_*(\mathcal{O}_X)=\mathcal{O_Y}$. When this condition arises has been discussed in particular in this MO question, and a related question is this one (there are, I think, some others too). My broad question is what are some interesting things we can deduce from this when it does occur?

My more specific question is about when $f_*$ sends locally free sheaves to locally free sheaves (which has been discussed here): If we know $f_*(\mathcal{O}_X)=\mathcal{O}_Y$, is there an easy condition which implies that $f_*$ takes vector bundles to vector bundles?

For example, it is clear if $f$ is open and surjective, or more generally if $Y$ has an open cover $\{V_i\}$ for which $f^{-1}(V_i)\subset X$ is contained in an open subscheme which admits only trivial bundles (e.g. $f^{-1}V_i\subset\mathbf{A}^n_k$).

I am happy to assume that $f$ is quasicompact and quasiseparated, or even that both $X$ and $Y$ are. The example I have in mind is the punctured plane $j\colon\mathbf{A}^2_k\setminus 0\hookrightarrow\mathbf{A}^2_k$. Since $\Gamma(\mathbf{A}^2_k,\mathcal{O_{\mathbf{A}^2_k}})=\Gamma(\mathbf{A}^2_k\setminus 0,\mathcal{O_{\mathbf{A}^2_k}})$, the qcqs lemma shows $j_*\mathcal{O_{\mathbf{A}^2_k\setminus0}}=\mathcal{O_{\mathbf{A}^2_k}}$. What might I show to conclude that $j_*\mathcal{E}$ is a vector bundle for any vector bundle $\mathcal{E}$, so in particular all vector bundles over $\mathbf{A}^2_k\setminus0$ are trivial.

Any help is appreciated!

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    $\begingroup$ Due, probably, to some quirk in the Markdown editor, you had two different links labelled [1]. I arbitrarily changed one of them to the first free number, [5], and updated what seemed to be the relevant link. I hope that was correct. $\endgroup$
    – LSpice
    Commented 3 hours ago
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    $\begingroup$ Note that the result would be false for $j\colon\mathbf{A}^n_k\setminus 0\hookrightarrow\mathbf{A}^n_k$ with $n\geq 3,$ so the "easy condition" you mention should be quite subtle. $\endgroup$
    – abx
    Commented 2 hours ago
  • $\begingroup$ @abx Indeed, that's true. I'm expecting some dimension hypotheses to show up such as in algebraic Hartog's lemma, or more closely related this result about coherent reflexive sheaves: stacks.math.columbia.edu/tag/0EBJ $\endgroup$
    – naahiv
    Commented 2 hours ago

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