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I'm struggling to understand this part of the proof of this proposition:

I'm having troubles only in this part, I really need help.

Thanks a lot.

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2 Answers

up vote 5 down vote accepted

Because $\varphi$ is a morphism of sheaves, the fact that $\varphi(V_P)(s(P)) = t|_{V_P}$ implies that $\varphi(V_P \cap V_Q)(s(P)|_{V_P \cap V_Q}) = t|_{V_P \cap V_Q}$. To be maximally explicit, this comes from commutativity of $$\begin{array}{ccc} \mathcal{F}(V_P) & \stackrel{\varphi(V_P)}{\longrightarrow } & \mathcal{G}(V_P) \\ \downarrow && \downarrow \\ \mathcal{F}(V_P \cap V_Q) &\stackrel{\varphi(V_P\cap V_Q)}{\longrightarrow } & \mathcal{G}(V_P \cap V_Q) \end{array}$$

where the vertical maps are restrictions. Look again at the definition of a sheaf morphism! Similarly $\varphi(V_P \cap V_Q)(s(Q)|_{V_P \cap V_Q}) = t|_{v_P \cap v_Q}$. Hence $\varphi(V_P \cap V_Q)$ sends $s(P)|_{V_P\cap V_Q}$ and $s(Q)|_{V_P\cap V_Q}$ to the same thing.

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But $\varphi=\varphi(U)$? what exactly is $\varphi$? –  user42912 Sep 17 '13 at 16:09
    
I'm confused, you use $\varphi$ to denote $\varphi (V_p)$ and $\varphi (V_P\cap V_Q)$ in the same time. –  user42912 Sep 17 '13 at 18:24
    
@user42912 Look again at the definition of $\varphi$; it is a collection of maps for every open set. –  fpqc Sep 18 '13 at 4:31
    
yes, yes I understood thank you very much for your help. –  user42912 Sep 18 '13 at 10:46
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The main point here is that morphisms of sheaves commute with restriction.

You have $\varphi(s(P)|_{V_P})=t|_{V_P}$, and $s(P)|_{V_P\cap V_Q}=(s(P)|_{V_P})|_{V_Q}$ (this is essentially one of the axioms of presheaves). So

$$\varphi(s(P)|_{V_P\cap V_Q})=\varphi(s|_{v_P})|_{V_Q}=(t|_{V_P})|_{V_Q}=t|_{V_P\cap V_Q}$$

Then you can do the same thing with $P$ and $Q$ interchanged to get $\varphi(s(Q)|_{V_P\cap V_Q})=t|_{V_P\cap V_Q}$.

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Minor nitpick in the second sentence: It's actually more generally one of the axioms of a presheaf. +1 Nonetheless. –  fpqc Sep 17 '13 at 15:20
    
@user38268 Ha, that is a fairly extreme level of pedantry! ;-) But it is indeed true that if you want to find the axiom I mean, you probably need to look up presheaves, so edited. –  Matt Pressland Sep 17 '13 at 15:27
    
What exactly is $\varphi$? can I say that $\varphi (U)=\varphi$? thank you. –  user42912 Sep 17 '13 at 16:45
    
@user42912 $\varphi$ is a collection of maps $\varphi(U)\colon\mathscr{F}(U)\to\mathscr{G}(U)$ for each open $U$, but the $(U)$ isn't always written to simplify the notation. All of the sections you consider are defined over open subsets of $U$, so everything makes sense when you replace $\varphi$ by $\varphi(U)$ again. –  Matt Pressland Sep 17 '13 at 17:59
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@user42912 For your last comment: Look at the axioms for a presheaf! –  fpqc Sep 18 '13 at 4:32
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