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 $W_i=\varphi^{-1}(V_i)$.