# [Definition included]Morphisms agree in intersections of a cover comes from a morphism $X\to Y$ for scheme $X,Y$

$X$ and $Y$ are schemes(Note they are not necessarily affine), and $(U_i)_{i∈I}$ is an affine open cover of $X$, and let $f_i: U_i → Y$ be morphisms which agree on the overlaps: for all $j,k ∈ I$, we have $f_j|_{U_j∩U_k} = f_k|_{U_j∩U_k}$. Prove there exists a unique morphism $f : X → Y$ such that for all $i ∈ I$, we have $f|_{U_i} = f_i$.

The following are the definitions we use here:

Definition of scheme: A functor $X$ is a scheme if it is (1)local;(2)has an open affine cover.

Definition of local: (Both of these two definitions are taken in Jantzen's text.)

Many thanks for any help. I have tried on this exercise from this morning almost without any progress(What I have tried so far was searching for a way to use the fact that the $U_i$'s are affine). So pointing out some useful key fact or just give an answer is very ideal.

• I'm not sure what you think the difficulty here is. The functors are local, so by definition maps are defined by their values on an open cover of the domain, which is exactly the data you are given (the $U_i$ and $f_i$). – Jim Nov 11 '17 at 13:05
• @Jim I tried to write down the sequence for $Hom(X,Y)$, so once I prove that the sequence is exact then I will solove the problem, but I am not able to prove it. Your argument sounds quite reasonable. I think from your idea I can use the given information to construct such a map. But how to prove that it is unique? – PropositionX Nov 12 '17 at 23:21
• You don't need to prove that the sequence is exact, you are told that $X$ and $Y$ are schemes, which by definition means that they're local. – Jim Nov 13 '17 at 0:12