Doubt in Hartshorne's book

Question

After to post this question: Help with a proof in Hartshorne's book I realized that I have another doubt in this proof:

What I know is

$\varphi_p(s_p)=t_p\implies (\varphi_{V_p}(s(P))_p=t_p$

I don't understand why $(\varphi_{V_p}(s(P))_p=(t_{|V_P})_p$

NOTATION:

Instead of $\varphi(V_P)$, to simplify notation I write $\varphi_{V_P}$

Sorry to post the same proof again

Thanks a lot.

Answer 1

Consider the morphisms

• $\rho_{X/V_P}:\mathcal{G}(X)\to\mathcal{G}(V_P)$
• $\sigma_{V_P/P}:\mathcal{G}(V_P) \to \mathcal{G}_P$
• $\sigma_{X/P}:\mathcal{G}(X)\to\mathcal{G}_P$

It follows from the definition of a stalk that $\sigma_{X/P}=\sigma_{V_P/P}\circ\rho_{X/V_P}$ because $V_P$ and $X$ are both open neighborhoods of $P$. Look up the definition of a direct limit to confirm this, if you like.

In other words, the above statement means $t_P=(t|_{V_P})_P$. Now given what you know, this yields what you want to understand.

Edit I would write it down as follows: We may replace $V_P$ by $V_P\cap U$, then $V_P$ is an open subset contained in both $V_P$ and $U$. Hence, we have $t_P=(t|_{V_P})_P$. Thus, $\varphi_P(s_P)=t_P$ implies $(\varphi_{V_P}(s(P)))_P=t_P=(t|_{V_P})_P$.

Answer 2

Lema 1

Let $V\subset U$ and $t\in \mathcal F(U)$, we have $(t_{|V})_p=t_p$

Proof

$(t_{|V})_p=t_p \Leftrightarrow \exists \ W\subset V\cap U=V;\ t_{|W}=(t_{|V})_{|W}$, but

$(t_{|V})_{|W}=t_{|W}$

Proof of the question:

$\varphi_P(s(P))=t_P\implies (\text{by definition of}\ \varphi_P)\\ (\varphi_{V_P}(s(P)))_P=t_P\implies\\ \exists\ W\subset V_P\cap U=V_P; (\varphi_{V_P}(s(P)))_{|W}=t_{|W}\implies \ (\text{since} \ t_{|W}=(t_{|V_P})_{|W})\\(\varphi_{V_P}(s(P)))_{|W}=(t_{|V_P})_{|W}\implies\\ ((\varphi_{V_P}(s(P)))_{|W})_P=((t_{|V_P})_{|W})_P\implies\text{(Lemma 1)}\\ (\varphi_{V_P}(s(P)))_P=((t_{|V_P}))_P$