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Exercise 1.8 in Hartshorne, Chapter II:

Let $0\to\mathcal{F}'\xrightarrow{\phi}\mathcal{F}\xrightarrow{\psi}\mathcal{F}''$ be a left-exact sequence. Then for any open set $U\subseteq X$, the sequence $0\to\Gamma(U,\mathcal{F}')\to\Gamma(U,\mathcal{F})\to\Gamma(U,\mathcal{F}'')$ is left-exact.

Here is the way I tried to prove this: Let $U\subseteq X$ be open. Since $\phi$ is injective, $\phi_U:\Gamma(U,\mathcal{F}')\to\Gamma(U,\mathcal{F})$ is injective. So it remains to show that $\ker(\psi_U)=\operatorname{im}(\phi_U)$. Then I looked at the diagram

$$\begin{array}{cccccc} 0 & \rightarrow & \Gamma(U,\mathcal{F}') & \xrightarrow{\phi_U} & \Gamma(U,\mathcal{F}) & \xrightarrow{\psi_U} & \Gamma(U,\mathcal{F}'')\\ & & \downarrow & & \downarrow & & \downarrow\\ 0 & \rightarrow & \mathcal{F}'_p & \xrightarrow{\phi_p} & \mathcal{F}_p & \xrightarrow{\psi_p} & \mathcal{F}^{''}_p \end{array}$$

which commutes for every $p\in U$ and with bottom row exact. Now if $s\in\Gamma (U,\mathcal{F}')$, we have $\psi_U(\phi_U(s))_p=\psi_p(\phi_p(s_p))=0$ for every $p\in U$, hence $\psi_U(\phi_U(s))=0$. The inclusion left to show is $\ker(\psi_U)\subseteq\operatorname{im}(\phi_U)$.

For this, take $t\in\ker(\psi_U)$. $\psi_U(t)=0$ implies $\psi_p(t_p)=\psi_U(t)_p=0$ for every $p$. Since $\ker(\psi_p)=\operatorname{im}(\phi_p)$, there exist $s'_p\in\mathcal{F}'_p$ such that $t_p=\phi_p(s'_p)$, hence also a section $s\in\Gamma(U,\mathcal{F}')$ with $s_p=s'_p$ (bad notation there, I didn't know how to name those germs). We have $\phi_U(s)_p=\phi_p(s_p)=t_p$ for every $p\in U$, so $\phi_U(s)=t\in\operatorname{im}(\phi_U)$.

Now my question is: is this proof correct? I usually look up some solutions in the internet after doing such an exercise, and for this one I found two solutions which both took a different approach from mine, which made me a bit insecure about my 'solution', since I took the 'direct approach' from my point of view.

Thank you in advance!

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Hello @Zhen Lin, that is very interesting, I did not know this. I'll read up on it, thank you! –  InvisiblePanda Feb 26 '12 at 11:49

1 Answer 1

up vote 2 down vote accepted

In my opinion, you need to reason why the $s'_P$ do glue to some $s'\in\mathcal{F}'(U)$. However, this is not that tough. We pick neighborhoods $V_P\subseteq U$ for each point $P$ such that $s'_P=(f_P,V_P)$ with $f_P\in\mathcal{F}'(V_P)$. For $W:=V_P\cap V_Q\ne 0$, we have $\phi_W(f_P|_W)=t|_W=\phi_W(f_Q|_W)$ and we know that $\phi_W$ is injective.

PS: That little nitpicking aside, I don't see anything wrong with the proof.

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Hello @rattle! I just suppressed that little step so my question wouldn't be that lenghty, since it's an argument that was used in the exercises before already ;) Thanks for proof-reading! –  InvisiblePanda Feb 26 '12 at 11:53

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