Help with a proof in Hartshorne's book

Question

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.

Answer 1

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.

Answer 2

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 $\varphi(s(Q)|_{V_P\cap V_Q})=t|_{V_P\cap V_Q}$.