[@Fa1 Theorem 11.2.9]) yields $$\xymatrix{0\ar[r] & H^0(B^n,W)\ar[r] & H^0(B^1,W)\ar[r] & H^1_{x_n}(B^n,W)\ar[r] & \ldots}$$ Using Lemma \[reg\_param\] and the computation of $H^i(B^1,W)$, we get an exact sequence $$\xymatrix{0\ar[r] & H^0(B^n,W)\ar[r] & \Z\oplus \Z\ar[r] & H^0(C^{n-2},W)\ar[r] & H^1(B^n,W)\ar[r] & 0}$$ and isomorphisms $H^{i-1}(C^{n-2},W)\simeq H^i(B^n,W)$ for $i\geq 2$. Since $C^0\simeq \C$, we have $H^{i-1}(C^0,W)=0$ for $i\geq 2$ and Lemma \[comput\_c1\] shows that the same statement is true for $C^1$. Hence $H^i(B^n,W)=0$ for $i\geq 2$ and it suffices to prove that the map $\Z\oplus \Z\to H^0(C^{n-2},W)$ in the above exact sequence is surjective to conclude. The two generators of $H^0(B^1,W)\simeq \Z\oplus \Z$ are the class of $\langle 1\rangle$ and $\langle x_n\rangle$. Now it is not hard to see that $\langle x_n\rangle$ is sent to the generator of $H^0(C^{n-2},W)$ (which is either $\Z/2$ or $\Z/4$) under the map $\Z\oplus \Z\to H^0(C^{n-2},W)$. Whence the result.

Next we compute part of the Witt cohomology of $B^3_{x_2+x_3}$. It is of course also possible to compute $H^0(B^3_{x_{2}+x_3},W)$, but the answer is not “nice” and we won’t need it.

\[comput\_b3\] We have $$H^i(B^3_{x_2+x_3},W)=\left\{ \begin{array}{cl}
\Z/2 & \mbox{if $i=1$,}\\
0 & \mbox{if $i\geq 2$.}
\end{array}\right.$$

Since we already know the Witt cohomology groups of $B^3$ (Lemma \[comput\_bn\]) and $B^3/(x_2+x_3)=C^2$ (Lemma \[comput\_c2\]), 