How I can show that a morphism of schemes $f:X\longrightarrow Y$ is separated if and only if for every open cover $(V_i)_{i\in I}$ of $Y$, the induced morphisms $f^{-1}(V_i)\longrightarrow V_i$ are separated?
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This is exactly Proposition (5.5.5) of EGA I. – QU Binggang Sep 24 '19 at 10:04
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Here's a general argument. – Elías Guisado Villalgordo Jun 12 '23 at 13:04
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Suppose $V_i$ is an open cover of $Y$, and suppose $U_i\to V_i$ is separated for each $i$ where $U_i:=f^{-1}(V_i)$. We want to show that the image of the diagonal map $\delta:X\to X\times_Y X$ is closed in $X\times_Y X$. To do this, it suffices to show that for some open cover $\mathcal U$, the image of $\delta^{-1}(O)\to O$ is closed in $O$ for each $O\in\mathcal U$. If you let $g:X\times_Y X\to Y$ be the natural map, then $g^{-1}(V_i)$ gives an open cover. Now, using definition of the fiber product, you should be able to rewrite $g^{-1}(V_i)$ in a nicer form, which by assumption will immediately imply that $\delta^{-1}(g^{-1}(V_i))\to g^{-1}(V_i)$ is closed for each $i$.
For the converse, maybe you are aware that separatedness is preserved by base change? If so then this is trivial, since the base change of $f:X\to Y$ by $V_i\hookrightarrow Y$ is just $f^{-1}(V_i)\to V_i$. If you didn't already know this, then it's not difficult to show, but if you're struggling with it then let me know and I can try to update my answer.
Alex Mathers
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