Proper Cech descent does not require acyclic fibers

The space and its two charts. Glue the affine $z$-line and affine $w$-line along $zw=1$. Their regular transition gives a prevariety with a finite affine cover ([F2], [F3]). Its diagonal is closed: on the two same-chart products equality is $z=z^{\mathrm{\prime }}$ or $w=w^{\mathrm{\prime }}$, and on either cross-chart product it is $z{w}^{\mathrm{\prime }}=1$ or $w{z}^{\mathrm{\prime }}=1$. Equalizers of regular maps are inverse images of this diagonal, so $S$ is separated. The formulas of [F5] define a homeomorphism from this glued space to $S^2$: in the $w$-chart they become $(2\mathrm{Re}w,-2\mathrm{Im}w,1-\mathrm{\mid }w{\mathrm{\mid }}^2)\mathrm{/}(1+\mathrm{\mid }w{\mathrm{\mid }}^2)$, including the north pole at $w=0$; the inverse coordinate there is $w=(x-iy)\mathrm{/}(1+t)$, continuous near the north pole. In the other chart the formulas and inverse are those of [F5]. Composing with [F5]'s inverse identifies the analytic topology with the one-point compactification of [F4]. In particular $S$ is compact.

The map is proper surjective. The unique morphism $f:S\to \mathrm{p}\mathrm{t}$ is surjective because $S$ is nonempty. It is projective in the sense of [F9]: the identity is a closed immersion $S\hookrightarrow \mathrm{p}\mathrm{t}\times {\mathbb{P}}^1(\mathbb{C})$ onto a reduced closed subvariety, and the projection to the point is the structure map. By the properness clause of [F10], a projective algebraic morphism is proper in the analytic topology; hence $f$ is a proper surjective morphism of finite-type complex algebraic varieties, and [F1] applies to it.

The single fiber is not acyclic. Stereographic projection is a homeomorphism $\widehat{\mathbb{C}}\to S^2$ ([F5]); a homeomorphism is a homotopy equivalence, so it induces isomorphisms on singular homology ([F6]) and $H_2(S;k)\cong H_2(S^2;k)\cong k$ by [F7]. The field-dual corollary [F8] then gives $H^2(S;k)\cong {\mathrm{Hom}}_k(H_2(S;k),k)\cong k\mathrm{\ne }0$, since $k$ is a field. The sphere is metric with a contractible-open neighbourhood basis, and these properties transfer through the homeomorphism, so [F11] identifies this singular cohomology group with the actual derived constant-sheaf group $H^2(S,k_S)$. Thus the fiber $S$ of $f$ is not cohomologically acyclic.

The rows are still contracted. The fiber of $f$ over the unique point is $S$ and the fiber powers are $S^{p+1}$. Choose one point $y\in S$, for instance the point with $z=0$, and for $r\ge 0$ let ${\mathrm{i}\mathrm{n}\mathrm{s}}_r:S^r\to S^{r+1}$ insert $y$ as first coordinate, with $S^0=\mathrm{p}\mathrm{t}$. For the compositions below the deletion maps are $d_i:S^{r+2}\to S^{r+1}$, and they satisfy $d_0{\mathrm{i}\mathrm{n}\mathrm{s}}_{r+1}=\mathrm{i}\mathrm{d}$ and $d_i{\mathrm{i}\mathrm{n}\mathrm{s}}_{r+1}={\mathrm{i}\mathrm{n}\mathrm{s}}_r{d}_{i-1}$ for $i>0$, so with $h_{p+1}:={\mathrm{i}\mathrm{n}\mathrm{s}}_{p+1}^{\ast }$ and ${\delta }_p$ the alternating sum of deletion pullbacks one gets $h_{p+1}{\delta }_p+{\delta }_{p-1}{h}_p=\mathrm{i}\mathrm{d}$ in every internal degree $q$, and $h_0{\delta }_{-1}=\mathrm{i}\mathrm{d}$; hence each augmented row is contractible. In the row $q=2$ the term at $p=0$ is $H^2(S;k)\cong k\mathrm{\ne }0$, carried injectively into $H^2(S^2;k)$; the nonzero group is retained and does not obstruct exactness.

Conclusion. Applying [F1] to $f$ and the constant sheaf $k$, and using that the fiber powers of $f$ are $S^{p+1}$ and that $R\mathrm{\Gamma }(\mathrm{p}\mathrm{t},k)=k$ is concentrated in degree $0$, the augmented Cech map $k\stackrel{\sim }{\to }{\mathrm{Tot}}_{p\ge 0}R\mathrm{\Gamma }(S^{p+1},k)$ is a quasi-isomorphism; therefore the unaugmented product total has cohomology $k$ in degree $0$ and zero in every other degree. The example shows that proper Cech descent does not require acyclic fibers: the nonvanishing $H^2$ of the fiber is retained throughout the contracted rows. The Axiom of Choice is inherited through the descent lemma and the field-dual corollary, and the construction of $S$, the homeomorphism with $S^2$, the choice of one point and the insertion argument use no further choice.