# Cech cohomology (Harthshorne)

I´m studying Cech cohomology through Hartshorne´s algebraic geometry and EGA, and I´m stuck in two exercises, I wonder if you can help me. The first one is the exercise III.4.4, we want to prove that $$\varinjlim_{\mathfrak{U}}\check{H}^{1}(\mathfrak{U},\mathcal{F})=H^{1}(X,\mathcal{F})$$ where $\mathfrak{U}$ is a covering of $X$. (I attach a photo of the exercise) Exercise III. 4.4

I have just proved $a)$ and $b)$.

I´m also stuck with the last exercise of this section: Let $X$ be a topological spaces, $\mathcal{F}$ a sheaf of abelian groups, and $\mathfrak{U}=(U_{i})$ an open cover. Assume for any finite intersection $V=U_{i_{0}}\cap\cdots\cap U_{i_{p}}$ of open sets of the covering, and for any $k>0$, that $H^{k}(V,\mathcal{F}_{\mid V})=0$. Then prove that for all $p\geq 0$, the natural maps $$\check{H}^{p}(\mathfrak{U},\mathcal{F})\rightarrow H^{p}(X,\mathcal{F})$$ of the last exercise (III. 4.4) are isomorphisms.

Thank you for your time.

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I follow the hint there and adapt the notations. For the statement (c), you can take the long exact sequence of the sequence of the complex $0\to C^\bullet(\mathfrak{U},\mathcal{F})\to C^\bullet(\mathfrak{U},\mathcal{G})\to D^\bullet(\mathfrak{U})\to 0$. This gives you

$$0\to \check{H}^0(\mathfrak{U},\mathcal{F})\to \check{H}^0(\mathfrak{U},\mathcal{G})\to H^0(D^\bullet(\mathfrak{U}))\to \check{H}^1(\mathfrak{U},\mathcal{F})\to 0.$$

$\check{H}^1(\mathfrak{U},\mathcal{G})=0$ since $\mathcal{G}$ is flasque.

Also you have isomorphisms $\check{H}^0(\mathfrak{U},\mathcal{F})\cong\Gamma(X,\mathcal{F})$ and $\check{H}^0(\mathfrak{U},\mathcal{G})\cong\Gamma(X,\mathcal{G})$ as well as a natural map $H^0(D^\bullet(\mathfrak{U}))\to \check{H}^0(\mathfrak{U},\mathcal{R})=\Gamma(X,\mathcal{R})$. These fit together and give the following commutative diagram:

$\require{AMScd}$ \begin{CD} 0@>>> \check{H}^0(\mathfrak{U},\mathcal{F})@>>> \check{H}^0(\mathfrak{U},\mathcal{G})@>>> H^0(D^\bullet(\mathfrak{U}))@>>> \check{H}^1(\mathfrak{U},\mathcal{F})@>>> 0\\ @. @VVV @VVV @VVV \\ 0@>>>\Gamma(X,\mathcal{F})@>>> \Gamma(X,\mathcal{G})@>>> \Gamma(X,\mathcal{R})@>>> H^1(X,\mathcal{F})@>>> 0 \end{CD}

So you have an induced map $\check{H}^1(\mathfrak{U},\mathcal{F})\to H^1(X,\mathcal{F})$. Now take the direct limit (with respect to the open cover) in the first line. The result will follow if we can show $$\varinjlim_{\mathfrak{U}} H^0(D^\bullet(\mathfrak{U}))\cong H^0(X,\mathcal{R}).$$ The injectivity is obvious. For surjectivity, from the definition of quotient sheaves, for any $r\in \Gamma(X,\mathcal{R})$, $r$ can be described by giving an open cover $\mathfrak{V}=\{V_i\}$ of $X$ and elements $g_i\in \Gamma(V_i,\mathcal{G})$ such that $g_i-g_j\in\Gamma(V_i\cap V_j,\mathcal{F})$. Hence the sujectivity follows.

For the last exercise, you can imitate the proof of Theorem 4.5. Instead of the affineness of the open sets, the condition on the vanishing of higher cohomologies in the exercise implies that you still have the short exact sequence $$0\to \mathcal{F}(U_{i_0\cdots i_p})\to \mathcal{G}(U_{i_0\cdots i_p})\to \mathcal{R}(U_{i_0\cdots i_p})\to 0$$ and hence the short exact sequence of Cech complexes. The same argument will work. You need the vanishing assumption here again for the induction step.

Prove that

$(*)$ under refinement $D^*(\mathfrak U) \to C^*(\mathfrak U, \mathscr R)$ induces an isomorphism on $h^0$.

Use that locally a section of $\mathscr R$ lifts to a section of $\mathscr G$.

Then you get an exact sequence, $$H^0(X, \mathscr G) \to H^0(X,\mathscr R) \to \check H^1(X,\mathscr F) \to 0.$$ Note $H^1(X,\mathscr F) = \text{coker}\left( H^0(X,\mathscr G) \to H^0(X,\mathscr R)\right)$.

• I dont have any idea about how to prove that – matemagreek Dec 29 '17 at 10:56
• To show (*) is a surjection on $h^0$ I just need to know that any section of $\mathscr R$ can locally be lifted to a section of $\mathscr G$. Do you think so? – Ben Dec 29 '17 at 17:12