Lemma 55.2.5. Let $L_0$, $L_1$ be a finite free $\mathbf{Z}$-modules endowed with integral symmetric bilinear positive definite forms $\langle \ ,\ \rangle : L_ i \times L_ i \to \mathbf{Z}$. Let $\text{d} : L_0 \to L_1$ and $\text{d}^* : L_1 \to L_0$ be adjoint. If $\langle \ ,\ \rangle $ on $L_0$ is unimodular, then there is an isomorphism
\[ \Phi : \mathop{\mathrm{Coker}}(\text{d}^*\text{d})_{torsion} \longrightarrow \mathop{\mathrm{Im}}(\text{d})^\# /\mathop{\mathrm{Im}}(\text{d}) \]
with notation as in Lemma 55.2.4.
Proof.
Let $x \in L_0$ be an element representing a torsion class in $\mathop{\mathrm{Coker}}(\text{d}^*\text{d})$. Then for some $a > 0$ we can write $ax = \text{d}^*\text{d}(y)$. For any $z \in \mathop{\mathrm{Im}}(\text{d})$, say $z = \text{d}(y')$, we have
\[ \langle (1/a)\text{d}(y), z \rangle = \langle (1/a)\text{d}(y), \text{d}(y') \rangle = \langle x, y' \rangle \in \mathbf{Z} \]
Hence $(1/a)\text{d}(y) \in \mathop{\mathrm{Im}}(\text{d})^\# $. We define $\Phi (x) = (1/a)\text{d}(y) \bmod \mathop{\mathrm{Im}}(\text{d})$. We omit the proof that $\Phi $ is well defined, additive, and injective.
To prove $\Phi $ is surjective, let $z \in \mathop{\mathrm{Im}}(\text{d})^\# $. Then $z$ defines a linear map $L_0 \to \mathbf{Z}$ by the rule $x \mapsto \langle z, \text{d}(x)\rangle $. Since the pairing on $L_0$ is unimodular by assumption we can find an $x' \in L_0$ with $\langle x', x \rangle = \langle z, \text{d}(x)\rangle $ for all $x \in L_0$. In particular, we see that $x'$ pairs to zero with $\mathop{\mathrm{Ker}}(\text{d})$. Since $\mathop{\mathrm{Im}}(\text{d}^*\text{d}) \otimes \mathbf{Q}$ is the orthogonal complement of $\mathop{\mathrm{Ker}}(\text{d}) \otimes \mathbf{Q}$ this means that $x'$ defines a torsion class in $\mathop{\mathrm{Coker}}(\text{d}^*\text{d})$. We claim that $\Phi (x') = z$. Namely, write $a x' = \text{d}^*\text{d}(y)$ for some $y \in L_0$ and $a > 0$. For any $x \in L_0$ we get
\[ \langle z, \text{d}(x)\rangle = \langle x', x \rangle = \langle (1/a)\text{d}^*\text{d}(y), x \rangle = \langle (1/a)\text{d}(y),\text{d}(x) \rangle \]
Hence $z = \Phi (x')$ and the proof is complete.
$\square$
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