y Theorem \[210\], there is a suitable basis for this lattice such that the norm of the $\pi^1$-modular Jordan component is the ideal $(4)$. Namely, we choose $$(e_5-e_1', e_1'-\frac{2\pi(b+b') }{\delta(1+4b')}e_2', \pi e_5+\frac{a}{1+4b'}e_2').$$ Here, a method to find the above basis follows from the argument used in Case (iii) of Case (1) with $j$ even. Then the lattice spanned by the latter two vectors is $\pi^1$-modular with the norm $(4)$ (so that it is isometric to $H(1)$ by Theorem \[210\]) and the lattice spanned by the first vector is $(a+4(b+b'))$. Let $$\left\{
    \begin{array}{l}
    \tilde{M}_0=\left(\oplus H(0)\right)\oplus B(e_5-e_1');\\
    \tilde{M}_1=\left(\oplus H(1)\right)\oplus \left( Be_3'\oplus Be_4'  \right) 
      \oplus \left( B(e_1'-\frac{2\pi(b+b') }{\delta(1+4b')}e_2')\oplus B(\pi e_5+\frac{a}{1+4b'}e_2')  \right).
    \end{array}\right.$$ Then $\tilde{M}_0\oplus\tilde{M}_1\oplus(\oplus_{i\geq 2}M_i)$ is another Jordan splitting of $L^{j-1}$, where $\tilde{M}_0$ is $\pi^0$-modular and *of type $I^o$* and $\tilde{M}_1$ is isometric to $\oplus H(1)$.

    For $\tilde{M}_0\oplus M_2$, the associated diagonal block of the image of a fixed element of $F_j$ in the special fiber of the smooth integral model associated to $L^{j-1}$ is $$\begin{pmatrix}id&0&0 \\ 0&1+2z_j&0
    \\ 0&0&id   \end{pmatrix}.$$ Here, the $(2,2)$-block corresponds to $B(e_5-e_1')$.

    We now follow the argument used in Step (3) in even case. Then we can easily check that the Dickson invariant of the image of a fixed element of $F_j$ in the orthogonal group associated to $M_0''$ is $(z_j)_1$.

    In conclusion, $(z_j)_1$ is the image of a fixed element of $F_j$ under the map $\psi_j$. Since $(z_j)_1$ can be either $0$ or $1$ by Equation (\[e42\]), $\psi_j|_{F_j}$ is surjective onto $\mathbb{Z}/2\mathbb{Z}$ and thus $\psi_j$ is surjective.\

So far, we have proved that $\psi_j$ is surjective. Let $\mathcal{B}$ be the set of integers $j$ such that $L_j$ is *of type I* and $L_{j+2}, L_{j+3}, L_{j+4}$ (resp. $L_{j-1}