is of the form $$\begin{gathered}
 \label{eq:3}
\left( \begin{matrix} B   & 0  \\ 0 &  C  \end{matrix} \right)\end{gathered}$$ with $B C^t = {\operatorname{id}}$. So $g$ is determined by its restriction $h$ to $\Lambda$ which is an arbitrary element of ${\operatorname{Gl}}( \Lambda )$.

Denote by $\pi$ the projection onto $E$ with kernel $\overline{E}$. Then the matrix of $ \pi g : E \rightarrow E$ in the basis $(e_i - i je_i)_i$ is $ \frac{1}{2} (  B + C)$. Furthermore $\det ( B + C)$ and $\det B$ have the same sign. So $\det ( \pi g : E \rightarrow E)$ is real and have the same sign of $\det (h)$. This proves the existence and unicity of $z'$.

To prove that $j_\Lambda$ is a morphism, we have to show that $ \zeta^{1/2} ( g'g E, g'E, E)$ is a positive number for any $g, g' \in {\operatorname{Sp}}( S, \Lambda,
j \Lambda)$. By the first part of the proof, $\zeta ( g'g E,
g'E, E)$ is real. So it suffices to prove it is positive for one pair $(h, h')$ in each component of ${\operatorname{Gl}}(\Lambda ) \times {\operatorname{Gl}}( \Lambda)$. If any two of three polarizations $E$, $F$ and $G$ are equal, then $$\zeta^{1/2} 