by $1\mapsto -ix_3Kos(x_1,x_2)$. Since $-i$ is a square, this is equivalent to the symmetric isomorphism $$\C(x_3)\to \exten 2{B^3_{(x_1,x_2)}}{\C(x_3)}{B^3_{(x_1,x_2)}}$$ defined by $1\mapsto x_3Kos(x_1,x_2)$. It is easy to check that the image of $(\kappa({\mathfrak q_2}), x_1x_3\rho_{x_2})$ is the symmetric isomorphism $$\C(x_3)\to \exten 2{B^3_{(x_1,x_2)}}{\C(x_3)}{B^3_{(x_1,x_2)}}$$ defined by $1\mapsto x_3Kos(x_2,x_1)$. Since $Kos(x_1,x_2)=-Kos(x_2,x_1)$, we see that $(\kappa({\mathfrak q_1}),(x_0x_1+x_2)\rho_{x_0^2+1})+(\kappa({\mathfrak q_2}), x_1x_3\rho_{x_2})$ indeed defines an element of the group $H^1(B^3_{x_2+x_3},W)$.

We next show that its image under the isomorphism $$H^1(B^3_{x_2+x_3},W)\to H^1(C^2,W)$$ is precisely the generator obtained in Remark \[explicit\_c2\]. We start with the computation of the image of $(\kappa({\mathfrak q_1}),(x_0x_1+x_2)\rho_{x_0^2+1})$. Arguing as above and using the equality $$(x_0x_1+x_2)(x_0x_1-x_2)=(x_0^2+1)(x_1^2+1)-x_2(x_2+x_3)$$ we see that $Kos(x_0^2+1,x_0x_1+x_2)=-x_2/(x_0x_1-x_2)Kos(x_0^2+1,x_2+x_3)$ in $\exten 2{B^3_{\mathfrak p_1}}{\kappa(\mathfrak p_1)}{B^3_{\mathfrak p_1}}$. Since $(x_0x_1-x_2)=-2x_2$ in $\kappa(\mathfrak p_1)$ (because $x_0x_1+x_2$ is in $\mathfrak p_1$) and $-1/2$ is a square, we see that the image of $(\kappa({\mathfrak q_1}),(x_0x_1+x_2)\rho_{x_0^2+1})$ is precisely the element $(\kappa(\mathfrak p_1),\rho_{x_0^2+1})$ defined in Remark \[explicit\_c2\]. It remains to compute the image of $(\kappa({\mathfrak q_2}), x_1x_3\rho_{x_2})$, but this is straightforward.

Witt cohomology of $S^2$
------------------------

Let $S^2=\R[x_0,x_1,x_2]/(\sum x_i^2-1)$. As in [@DF], we use the stereographic projection to compute the Witt cohomology of $S^2$ (and therefore its Witt groups). Explicitly, consider the algebra $\R[x_0,x_1]$ and the polynomial $f=x_0^2+x_1^2+1$. Then there is an isomorphism 