The singular cohomology with integer coefficients of a projective variety is isomorphic to the Čech cohomology of the constant sheaf of integers on this variety. If the above statement is correct then consider the following example: Look at $\mathbb{P}^1$ and the standard open cover $U_0,U_1$. Then the Čech complex for any sheaf $\mathcal{F}$ on $\mathbb{P}^1$ would be $$ 0 \rightarrow \mathcal{F}(U_0)\oplus\mathcal{F}(U_1) \rightarrow \mathcal{F}(U_0\cap U_1) \rightarrow 0. $$ If we consider the constant sheaf of integers this becomes $$ 0 \rightarrow \mathbb{Z}\oplus\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0 $$ with non-trivial map $(a,b)\mapsto a-b$. But the singular cohomology of $\mathbb{P}^1$ is $$H^i\mathbb{P}^1=\begin{cases} \mathbb{Z},\quad i=0,2 \\ 0,\quad \text{else.}\end{cases}$$ Where am I going wrong?
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7$\begingroup$ The comparison between singular and Cech cohomology only holds if you use Cech cohomology on the variety equipped with the analytic topology, which differs from the one using Zariski topology. $\endgroup$– Wojowu2 days ago
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$\begingroup$ @Wojowu Which of the above statements would not hold if we would fix the analytical topology? I can't see where Zariski topology is required. $\endgroup$– thrn2 days ago
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5$\begingroup$ In the definition of Čech cohomology (for analytic topology), you should take the (filtered) colimit over all coverings, not a single one. $\endgroup$– Z. M2 days ago
1 Answer
The Cech complex only computes cohomology if the subspaces have vanishing cohomology themselves: you need $$ H^i(U_0; \mathcal{F}) \cong H^i(U_1; \mathcal{F}) \cong H^i(U_0 \cap U_1; \mathcal{F}) \cong 0 $$ for all $i > 0$. In the example you've written, this is not true for $U_0 \cap U_1$. In general there is actually a Mayer-Vietoris long exact sequence relating the cohomology of $X$, $U_0$, $U_1$, and $U_0 \cap U_1$, or more generally a Mayer-Vietoris spectral sequence if the cover has more open sets.
(As a remark, you may have seen this constraint pushed under the rug because it is automatically true in special cases. If $X$ is a scheme and the subspaces $U_0$, $U_1$, and $U_0 \cap U_1$ are affine---the last automatic if $X$ is separated---then Serre's vanishing theorem says that these higher cohomologies vanish whenever $\mathcal{F}$ is a quasicoherent sheaf of $\mathcal{O}_X$-modules.)
