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I want to prove that the property of $f:X\to Y$ "universally close" is local on the base $Y$ ie if $Y=\cup_i V_i$ with for all $i$, $f^{-1}V_i\to V_i$ universally close then $f:X\to Y$ is universally close ie $\forall \varphi:Y'\to Y$, $X\times_Y Y'\to Y'$ is close.

Ideas: close is a property local on the base so it suffice to have $X\times_Y W_i\to W_i$ close for $W_i$ cover $Y'$. As $f^{-1}V_i\to V_i$ is universaly close then $f^{-1}V_i\times_{V_i}W_i\to W_i$ is a close morphism. But how then show that $X\times_Y W_i\to W_i$ is close?? For the $W_i$ we could take $W_i=\varphi^{-1}(V_i)$.

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Because of the way fiber products of schemes are built from affine opens, it is enough to prove that closedness is local on the base. This should be true for arbitrary maps of topological spaces: if $f:X\rightarrow Y$ is a map of topological spaces and there is an open cover $Y=\bigcup_iV_i$ with $f^{-1}(V_i)\rightarrow V_i$ closed, then $f$ is closed. – Keenan Kidwell Dec 16 '12 at 14:35
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As you take $W_i=\varphi^{-1}(V_i)$, $\varphi$ maps $W_i$ to $V_i$. So $$X\times_Y W_i=(X\times_Y V_i)\times_{V_i}\times W_i=f^{-1}(V_i)\times_{V_i} W_i.$$

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Ok, if I understand well, with notation $f_{Y'}:X\times_Y Y'\to Y'$ and $p:X\times_Y Y'\to X$ we have $X\times_Y W_i=f_{Y'}^{-1}(W_i)=f_{Y'}^{-1}(\varphi^{-1}(V_i))=p^{-1}f^{-1}(V_i)=f^{-1}(V_i‌​)\times_{V_i}\varphi^{-1}(V_i)=f^{-1}(V_i) \times_{V_i} W_i$ – Macadam Dec 17 '12 at 14:27
Yes, I edited my answer to add an intermediate step. – user18119 Dec 17 '12 at 14:37
Still ok, I detail the last equality for slowness people like me: $X\times_Y V_i\simeq q^{-1}(V_i)\subseteq X\times_Y Y$ with $q:X\times_Y Y\to Y$ but $X\times_Y Y\simeq(X,\mathrm{Id}_X,f)$ so $X\times_Y V_i\simeq f^{-1}(V_i)$ – Macadam Dec 17 '12 at 15:33

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