**Proof.**
We will use the properties of the functor $M \mapsto M^\vee $ found in More on Algebra, Section 15.55 without further mention. Part (1) holds because $\mathop{\mathrm{Hom}}\nolimits _ A(N, M^\vee ) = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N \otimes _ A M, \mathbf{Q}/\mathbf{Z})$ and because $\mathbf{Q}/\mathbf{Z}$ is injective in the category of abelian groups. Hence if $I \to (I^\vee )^\vee $ is split, then $I$ is pure injective. We claim that for any $A$-module $M$ the evaluation map $ev : M \to (M^\vee )^\vee $ is universally injective. To see this note that $ev^\vee : ((M^\vee )^\vee )^\vee \to M^\vee $ has a right inverse, namely $ev' : M^\vee \to ((M^\vee )^\vee )^\vee $. Then for any $A$-module $N$ applying the exact faithful functor ${}^\vee $ to the map $N \otimes _ A M \to N \otimes _ A (M^\vee )^\vee $ gives

\[ \mathop{\mathrm{Hom}}\nolimits _ A(N, ((M^\vee )^\vee )^\vee ) = \Big(N \otimes _ A (M^\vee )^\vee \Big)^\vee \to \Big(N \otimes _ A M\Big)^\vee = \mathop{\mathrm{Hom}}\nolimits _ A(N, M^\vee ) \]

which is surjective by the existence of the right inverse. The claim follows. The claim implies (3) and the necessity of the condition in (2).
$\square$

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