Failure of singular compactness for Hom

Assuming Gödel’s axiom of constructibility $V=L$, we construct a $\chi$-free abelian group $G$ of singular cardinality for some suitable cardinal $\chi$ which is regular and uncountable, equipped with the property that for every nontrivial subgroup $G^{\mathrm{\prime }}\subseteq G$ of smaller cardinality, $Hom(G^{\mathrm{\prime }},\mathbb{Z})\mathrm{\ne }0$, while $Hom(G,\mathbb{Z})=0$. This provides a consistent counterexample to the singular compactness of nontrivial duality with respect to the functor $Hom(-,\mathbb{Z})$.