$\gamma^w \oplus \gamma^{f_m+w}$ is a trivial bundle over $B_{m-1}$. Hence $P({\mathbb{C}}\oplus \gamma^{f_m})=P(\gamma^w \oplus \gamma^{f_m+w})=B_{m-1}\times {\mathbb{C}P}^1$. So we can reduce the twist number of $B_n$ to $t-1$, which is a contradiction.

**We now prove the claim.** Since $\psi$ is an isomorphism, we can write $$y_i = \sum_{j=1}^n b_{ij} \psi(x_j).$$ Let $B = (b_{ij})$ be the coefficient matrix. Note that $\det(B) = \pm1$.

Since $\psi^{-1}(y_k^2) = 0$ in $H^\ast(B_n)$ for $1 \leq k \leq n-s$, we have $$\begin{aligned}
    \psi^{-1}(y_k^2) &= (\psi^{-1}(y_k))^2 = (\sum_{j=1}^n b_{kj}x_j)^2 \nonumber \\
    & = \sum_{j=1}^n (b_{kj})^2x_j^2 + \sum_{1\leq i<j\leq n} 2b_{kj}b_{ki}x_ix_j \label{eqn:6}\\
    &= \sum_{j=1}^{n}(b_{kj})^2(x_j^2 - f_jx_j) \text{, which represents zero in $H^\ast(B_n)$}  \label{eqn:7} \\
    &= \sum_{j=1}^{n}(b_{kj})^2(x_j^2 - \sum_{i=1}^{j-1}\c{j}{i}x_ix_j)\nonumber\end{aligned}$$ By comparing the coefficients of and , we have $$\sum_{i=1}^{j-1} 2b_{kj}b_{ki}x_i =
    -(b_{kj})^2f_j \label{eqn:8}$$ for $ 1\leq k \leq n-s$ and $1\leq j \leq n$. This implies $$\label{eqn:9}
    2b_{kj}b_{ki} = -(b_{kj})^2 \c{j}{i}$$ where $ 1\leq k \leq n-s$ and $ 1\leq i <
j$.

Suppose that all $b_{kj}$ are even for $ n-t+1 \leq j \leq n$ and $ 1\leq k \leq n-s$. Since $t + n - s > n$, $\det B$ must be even because, in general, if $A = \left(
\begin{array}{cc}
C & D \\
E & 