Separated morphism of schemes (local on the target)

Question

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?

Answer 1

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.