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Let $\otimes$ denotes the usual tensor products of complexes and symbols live in the category of chain complexes of $R$-modules. Let $X$ be a dg-flat complex (i.e. $X_n$ is flat for each n and $X\otimes E$ is exact for each exact complex E). We can define a new complex $X^+={\rm Hom}_Z(X, \mathbb{Q}/\mathbb{Z})$. Is it true to say that $X^+$ is a dg-injective chain complex?

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I take 'dg-injective' as fibrant in the injective model structure on complexes of right dg-$R$-modules, whose weak equivalences are the quasi-isomorphisms and whose cofibrations are the levelwise injections ($X$ is therefore a complex of left $R$-modules). Then the answer is yes. I'll prove it satisfies the required lifting property. Consider a commutative square of complexes of right dg-$R$-modules $$\begin{array}{ccc} A&\rightarrow &\hom_{\mathbb Z}(X,\mathbb Q/\mathbb Z)\\ \downarrow&&\downarrow\\ B&\rightarrow&0 \end{array} $$ where $i\colon A\stackrel{\sim}\rightarrowtail B$ is a quasi-isomorphism which is also a levelwise inclusion. The adjoint of this diagram is a commutative square of complexes of abelian groups $$\begin{array}{ccc} A\otimes_R X&\rightarrow &\mathbb Q/\mathbb Z\\ \downarrow&&\downarrow\\ B\otimes_R X&\rightarrow&0 \end{array} $$ Here $i\otimes_R X\colon A\otimes_R X\rightarrow B\otimes_R X$ is levelwise injective since $X$ is levelwise flat and direct sums in the category of abelian groups are exact. Moreover, the cokernel of $i\otimes_RX$ is $(B/A)\otimes_R X$, which is exact since $B/A$ is exact (recall that $i$ is a an injective quasi-isomorphism). Therefore $i\otimes_RX$ is a cofibration in the injective model structure on chain complexes of abelian groups. Hence the latter square has a lifting $B\otimes_RX\rightarrow \mathbb Q/\mathbb Z$. The adjoint map $B\rightarrow \hom_{\mathbb Z}(X,\mathbb Q/\mathbb Z)$ is a lifting of the former. This concludes the argument.

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