F \\
\end{array}
\right)$ is an $n \times n$ matrix and if $D$ is a $k \times \ell$ matrix all of whose entries are even with $k + \ell >n$, then $\det B$ is even. This is a contradiction. Thus there is an odd number $b_{\ell m}$ for some $ 1 \leq \ell \leq n-s$ and $ n-t+1 \leq m \leq n$.

Suppose $f_m$ is not congruent to 0 modulo 2, i.e., there exists an odd number $\c{m}{h}$ for some $1\leq h \leq m-1$. Then from , $2b_{km}b_{kh} = -(b_{km})^2 \c{m}{h}$ for $1 \leq k \leq n-s$. It implies that $b_{km} \equiv 0 \text{ (mod $2$)}$ for all $1 \
\leq k \leq n-s$, which contradicts to that $b_{\ell m}$ is odd. Thus, $f_m \equiv 0 \text{ (mod $2$)}$.

On the other hand, from , $\sum_{i=1}^{m-1} 2b_{km}b_{ki}x_i = -(b_{km})^2f_m$ with $k=\ell$, $\frac{f_m}{2} = - \sum_{j=1}^{m-1} \frac{b_{\ell j}}{b_{\ell m}}x_j$. Thus we have $$\begin{aligned}
    \left( \frac{f_m}{2} \right)^2 &= \left(- \sum_{j=1}^{m-1} \frac{b_{\ell j}}{b_{\ell m}}x_j\right)^2 \\
    &= \sum_{j=1}^{m-1} \left(\left(\frac{b_{\ell j}}{b_{\ell m}}\right)^2 x_j^2 +2
 \sum_{h=1}^{j-1} \frac{b_{\ell j}b_{\ell h}}{(b_{\ell m})^2}x_j x_h \right) \\
    &= \frac{\sum_{j=1}^{m-1} (b_{\ell j})^2 (x_j^2 - f_j x_j)}{(b_{\ell m})^2} \hspace{2cm} \text{by \eqref{eqn:8}}\\
    &=0 \in H^\ast(B_{n-1})\end{aligned}$$

This proves the claim.

From the proof of 