Let $F$ be a number field (finite extension of $\mathbb{Q}$) and $\mathcal{O}_F$ its ring of integers. Let $I$ be a fractional ideal of $F$, and let $C=\{x \in F \mid \operatorname{Tr}(xy) \in \mathbb{Z},\ \forall y \in \mathcal{O}_F \}$ be the codifferent of $F/\mathbb{Q}$. Some thing I'm reading claims we have an isomorphism (I guess of $\mathbb{Z}$-modules): $$\operatorname{Hom}_{\mathbb{Z}}(I, \mathbb{Z}) \cong I^{-1} \otimes C $$ via the trace pairing:
Given an element $x \otimes c \in I^{-1} \otimes C$ with $x \in I^{-1}, c \in C$, we can get a map $I \to \mathbb{Z}$ sending $i \mapsto \operatorname{Tr}(ixc)$, note that $\operatorname{Tr}(ix \cdot c) \in \mathbb{Z}$ since $ix \in \mathcal{O}_F= II^{-1}$. This induces a map $I^{-1}\otimes C \to \operatorname{Hom}(\mathbb{Z}, I)$
Anyways, my question why is this an isomorphism? (For example I'm not sure how to construct the map in the reverse direction).