# extend a morphism knowing it on etale covering

Let $S$ be a separated scheme. Let $U,J$ be separated schemes over $S$. Assume we can construct, after etale base change $T\rightarrow S$, a map $U_T\rightarrow J_T$ (the sub-T indicates the pulled back scheme under base change). The question is why by descend theory (??) I get a map $U\rightarrow J$ ?

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In general, you don't. The two pull-backs to schemes over $T \times_S T$ must coincide: this is basically the statement that schemes are sheaves in the etale topology (even fpqc). –  Akhil Mathew Jul 28 '11 at 14:54
So to be more precise, let $X, Y$ be schemes. Let $Y' \to Y$ be a fpqc morphism. Let $Y' = Y' \times_Y Y'$, and let $X', X''$ be the appropriate base-changes. The statement is that $\hom(X, Y) \to \hom(X', Y') \rightrightarrows \hom(X'', Y'')$ is exact as a sequence of sets. –  Akhil Mathew Jul 28 '11 at 14:55
@Akhil Mathew I guess the left map should be injective. Is the sheaf in question $hom(X_T,Y_T)$? It is not a scheme, nor represented by one in general. I think what you are saying is correct for the sheaf $F(X)=hom(X,Y)$ because in Zariski topology we have that to give such a map is enough to give it on a covering and then check compatibility on the intersections. On the other hand I understand that you wrote down is correct in Zariski topology, but why in the etale or fpqc? could you give me references? –  unkn22 Jul 28 '11 at 15:17
The fact that schemes are sheaves in the fpqc topology is proved in, say, Angelo Vistoli's notes on descent theory (arxiv.org/abs/math/0412512) and SGA1 for much more. The key point it boils down to is that if $A$ is a ring and $B$ a faithfully flat $A$-algebra, then $A \to B \rightrightarrows B^{\otimes 2}$ is an exact sequence (which is proved by first assuming $A \to B$ has a section -- in which case one defines a retraction directly -- and reducing to this case by faithfully flat base change $A \to B$). –  Akhil Mathew Jul 29 '11 at 1:54