Let $h\colon S' \rightarrow S$ be a faithfully flat quasi-compact morphism. Let $f, g\colon X \rightarrow Y$ be morphisms of $S$-schemes. Let $X', Y'$, and $f', g'$ be the base changes by $h$. Suppose $f' = g'$. Can we say $f = g$?
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Yes, and as with all your other recent questions, you can find the answer in EGA, SGA, textbooks such as the one by Görtz-Wedhorn, etc.. You should also have a look at FGA and in particular at Vistoli's notes on descent theory. It is a waste of time to ask these questions because they are all answered in the standard references.
If $f',g' : X_{S'} \to Y_{S'}$ are equal, then so are $f \circ p, g \circ p : X_{S'} \to X \to Y$. Since $p : X_{S'} \to X$ is faithfully flat and quasi-compact, hence an (effective) epimorphism by descent theory, it follows $f=g$.
Martin Brandenburg
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[It is a waste of time to ask these questions because they are all answered in the standard references.] Is there an online reference written in English? I don't have an easy access to a university library. – Makoto Kato Dec 27 '13 at 19:00
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The stacks project for example. And Vistoli's descent theory notes. – Martin Brandenburg Dec 27 '13 at 19:09
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I checked the both. There's not the proof in Vistoli's note. I could not find the proof in the stacks project. Could you tell me the page or proposition no.? – Makoto Kato Dec 27 '13 at 19:17
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You certainly find that fpqc morphisms are effective epis and this is obviously equivalent to your question. – Martin Brandenburg Dec 27 '13 at 19:27
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What about preservation of various properties of morphisms by fpqc base changes? – Makoto Kato Dec 27 '13 at 19:46
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3What about reading any of the references which I have mentioned?? – Martin Brandenburg Dec 27 '13 at 19:48
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There is not the proof in Vistoli's note. I could not find the proof in the stack's project. Could you tell me the page or proposition no.? – Makoto Kato Dec 27 '13 at 19:52