Lemma 55.4.4. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. Assume $n$ is a $(-1)$-index. Let $T'$ be the numerical type constructed in Lemma 55.3.9. There exists an injective map
whose cokernel is an elementary abelian $2$-group.
Proof. Recall that $n' = n - 1$. Let $e_ i$, resp., $e'_ i$ be the $i$th basis vector of $\mathbf{Z}^{\oplus n}$, resp. $\mathbf{Z}^{\oplus n - 1}$. First we denote
and we set
A computation (which we omit) shows there is a commutative diagram
Since the cokernel of the top arrow is $\mathop{\mathrm{Pic}}\nolimits (T)$ and the cokernel of the bottom arrow is $\mathop{\mathrm{Pic}}\nolimits (T')$, we obtain the desired homomorphism of Picard groups. Since $\frac{w_ i}{w'_ i} \in \{ 1, 2\} $ we see that the cokernel of $\mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (T')$ is annihilated by $2$ (because $2e'_ i$ is in the image of $p$ for all $i \leq n - 1$). Finally, we show $\mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (T')$ is injective. Let $L = (l_1, \ldots , l_ n)$ be a representative of an element of $\mathop{\mathrm{Pic}}\nolimits (T)$ mapping to zero in $\mathop{\mathrm{Pic}}\nolimits (T')$. Since $q$ is surjective, a diagram chase shows that we can assume $L$ is in the kernel of $p$. This means that $l_ na_{ni}/w'_ i + l_ iw_ i/w'_ i = 0$, i.e., $l_ i = - a_{ni}/w_ i l_ n$. Thus $L$ is the image of $-l_ ne_ n$ under the map $(a_{ij}/w_ j)$ and the lemma is proved. $\square$
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