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Suppose we have a free abelian group $G=\mathbb{Z}^{n}$ and suppose we have an other abelian group $K$ such that there exists an injective homomorphism of groups $G\rightarrow Hom_{\mathbb{Z}}(K,\mathbb{Z})$. If we take the dual again

$$ Hom_{\mathbb{Z}}(Hom_{\mathbb{Z}}(K,\mathbb{Z}),\mathbb{Z}) \rightarrow Hom_{\mathbb{Z}}(G,\mathbb{Z})=G $$ is the evaluation map $$ Ev:K\rightarrow Hom_{\mathbb{Z}}(Hom_{\mathbb{Z}}(K,\mathbb{Z}),\mathbb{Z}) \rightarrow G$$ surjective ?

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The map $Hom(Hom(K,\mathbb{Z}),\mathbb{Z}) \to Hom(G,\mathbb{Z})$ need not be surjective, so its composition with $K \to Hom(Hom(K,\mathbb{Z}),\mathbb{Z})$ also need not be surjective.

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