Lemma 15.55.7. For any $R$-module $M$ the evaluation map $ev : M \to (M^\vee )^\vee $ is injective.

**Proof.**
You can check this using that $\mathbf{Q}/\mathbf{Z}$ is an injective abelian group. Namely, if $x \in M$ is not zero, then let $M' \subset M$ be the cyclic group it generates. There exists a nonzero map $M' \to \mathbf{Q}/\mathbf{Z}$ which necessarily does not annihilate $x$. This extends to a map $\varphi : M \to \mathbf{Q}/\mathbf{Z}$ and then $ev(x)(\varphi ) = \varphi (x) \not= 0$.
$\square$

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