$\mathrm{Sp}(B_i/Y_i, h_i)$).}$$ This equation will be proved in Appendix \[App:AppendixA\]. Thus $\mathrm{Im~}\varphi$ contains the identity component of $\prod_{i:even} \mathrm{O}(B_i/Z_i, \bar{q}_i)_{\mathrm{red}}\times \prod_{i:odd} \mathrm{Sp}(B_i/Y_i, h_i)$. Here $\mathrm{Ker~}\varphi$ denotes the kernel of $\varphi$ and $\mathrm{Im~}\varphi$ denotes the image of $\varphi.$ Note that it is well known that the image of a homomorphism of algebraic groups is a closed subgroup.

Recall from Section \[m\] that a matrix form of an element of $\tilde{G}(R)$ for a $\kappa$-algebra $R$ is written $(m_{i,j}, s_i \cdots w_i)$ with the formal matrix interpretation $$m= \begin{pmatrix} \pi^{max\{0,j-i\}}m_{i,j} \end{pmatrix}  \textit{ together with } z_i^{\ast}, m_{i,i}^{\ast}, m_{i,i}^{\ast\ast}.$$ We represent the given hermitian form $h$ by a hermitian matrix $\begin{pmatrix} \pi^{i}\cdot h_i\end{pmatrix}$ with $\pi^{i}\cdot h_i$ for the $(i,i)$-block and $0$ for the remaining blocks, as in Remark \[r33\].(1).

Let $\mathcal{H}$ be the set of even integers $i$ such that $\mathrm{O}(B_i/Z_i, \bar{q_i})_{\mathrm{red}}$ is disconnected. Notice that $\mathrm{O}(B_i/Z_i, \bar{q_i})_{\mathrm{red}}$ is disconnected exactly when $L_i$ with $i$ even is *free of type II*. We first prove that $ \varphi_i$, for such an even integer $i$, is surjective. We prove this by a series of reductions, after which we will be able to assume that $L$ is of rank two.

For such an even integer $i$ with a *free of type II* lattice $L_i$, we define the closed scheme $H_i$ of $\tilde{G}$ by the equations $m_{j,k}=0$ if $ j\neq k$, and $m_{j,j}=\mathrm{id}$ if $j \neq i$. An element of $H_i(R)$ for a $\kappa$-algebra $R$ can be represented by a matrix of the form $$\begin{pmatrix} id&0& & \ldots& & &0\\ 0&\ddots&& & & &\\ & &id& & & & \\  \vdots & & &m_{i,i} & & &\vdots
\\ & & & & id & & \\  & & & & &\ddots &0 \\ 0& & &\ldots & &0 &id \end{pmatrix}
\textit{ with $z_j^{\ast}=0, m_{j,j}^{\ast}=0, m_{j,j}^{\ast\ast}=0.$}$$ Obviously, $H_i$ has a grou