# An example that Čech cohomology is not equal to derived cohomology with on an affine scheme with Zariski topology.

There is an example in 3.1.10 $$\mathbb{A}^1$$-homotopy theory of schemes demonstrating that the Čech cohomology on an affine scheme with Zariski topology can be different from the derived cohomology.

Let $$x_1,x_2 \in \mathbb{A}^2$$ be two distinct closed points and let $$S$$ be the semilocal ring at $$x_1$$ and $$x_2$$. Let $$C_1,C_2 \subset \mathbb{A}^2$$ be two irreducible curves intersecting transversely at $$x_1,x_2$$. Let $$U= \mathrm{Spec}\, S\backslash(C_1\cup C_2), V = \mathrm{Spec}\, S \backslash\{x_1,x_2\}$$. Let $$j$$ be the inclusion $$U \rightarrow \mathrm{Spec}\, S$$.

We have the decomposition $$V= (V\backslash(V\cap C_1))\cup (V\backslash(V\cap C_2))$$ whose Mayer-Vietoris sequence give $$0=H^0(V\backslash(V\cap C_1),j_!\mathbb{Z})\oplus H^0(V\backslash(V\cap C_2),j_!\mathbb{Z})\rightarrow H^0(U,j_!\mathbb{Z}) = H^0(U,\mathbb{Z})=\mathbb{Z}\rightarrow H^1(V,j_!\mathbb{Z}).$$ Therefore, $$H^1(V,j_!\mathbb{Z})\neq 0$$. The decomposition $$\mathrm{Spec}\, S = (\mathrm{Spec}\, S\backslash x_1 )\cup (\mathrm{Spec}\, S\backslash x_2 )$$ gives a Mayer-Vietoris sequence $$H^1(\mathrm{Spec}\, S\backslash x_1,j_!\mathbb{Z})\oplus H^1(\mathrm{Spec}\, S\backslash x_2,j_!\mathbb{Z})\rightarrow H^1(V,j_!\mathbb{Z}) \rightarrow H^2(S,j_!\mathbb{Z}).$$

Then he conclude that $$H^2(S,j_!\mathbb{Z})\neq 0$$. But how can he says that the image of the nonzero element in $$H^1(V,j_!\mathbb{Z})\neq 0$$ is not zero in $$H^2(S,j_!\mathbb{Z})$$? (He says that this is because $$C_1$$ and $$C_2$$ are irreducible. But I don't know how this can leads to the conclusion.)

Write $$S$$ for the spectrum of the semi-local ring our considering, $$S_i := S \setminus \{x_i\}$$, and $$D_i := S_i \cap (C_1\cup C_2) = S_i \setminus U$$. Then, I think $$H^1(S_i,j_!\mathbb{Z}) = 0$$. So the morphism $$H^1(V,j_!\mathbb{Z}) \to H^2(S,j_!\mathbb{Z})$$ is injective.
I would suggest you to consider the following exact sequence on $$S_i$$: $$0 \to j_! \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}|_{D_i} \to 0.$$ Since $$\mathbb{Z}$$ is a constant sheaf, and any open subset of $$S_i$$ is irreducible (in particular, connected), $$\mathbb{Z}$$ is flasque. Hence $$H^l(S_i,\mathbb{Z}) = 0$$ for any $$l \geq 1$$, and $$H^0(S_i,\mathbb{Z}) = \mathbb{Z}$$. Moreover, since $$D_i$$ is connected, $$H^0(S_i,\mathbb{Z}|_{D_i}) = H^0(D_i,\mathbb{Z}) = \mathbb{Z}$$. These imply that $$H^1(S_i,j_!\mathbb{Z}) = 0$$ (note that the morphism $$\mathbb{Z} = H^0(S_i,\mathbb{Z}) \to H^0(S_i,\mathbb{Z}|_{D_i}) = \mathbb{Z}$$ is induced by $$\mathbb{Z} \to \mathbb{Z}|_{D_i}$$, hence this is an isomorphism): $$[H^0(S_i,\mathbb{Z}) = \mathbb{Z}] \xrightarrow{1\mapsto 1} [H^0(S_i,\mathbb{Z}|_{D_i}) = \mathbb{Z}] \to H^1(S_i,j_!\mathbb{Z}) \to [H^1(S_i,\mathbb{Z}) = 0].$$