Given two spectra $A$ and $B$, the set $[A,B]$ of homotopy classes of maps from $A$ to $B$ forms an abelian group. Can the dual abelian group $\text{Hom}([A,B],\mathbb{Q}/\mathbb{Z})$ be expressed as a group of homotopy classes of maps between spectra?


The *Brown-Comenetz dual* $I\_{\mathbb{Q}/\mathbb{Z}}E$ of a spectrum $E$ is defined [here](http://math.mit.edu/~vstojanoska/talk3.pdf). A key property of $I\_{\mathbb{Q}/\mathbb{Z}}E$ is that its homotopy is dual of that of $E$ in the sense that $\pi\_\*I\_{\mathbb{Q}/\mathbb{Z}}E=\text{Hom}(\pi\_\*E,\mathbb{Q}/\mathbb{Z})$.


Naively, I suspect that $\text{Hom}([A,B],\mathbb{Q}/\mathbb{Z})$ can be expressed in terms of $I\_{\mathbb{Q}/\mathbb{Z}}A$ and $I\_{\mathbb{Q}/\mathbb{Z}}B$.


What if I replace ${\mathbb{Q}/\mathbb{Z}}$ by ${\mathbb{R}/\mathbb{Z}}$ or ${\mathbb{C}/\mathbb{Z}}$?



pontryagin dual and maps between spectra
I think this is the closest thing there is to what you are looking for. Let $G$ be any injective (i.e., divisible) abelian group. Then there is a spectrum $I\_G$ representing the cohomology theory $X\mapsto \operatorname{Hom}(\pi^s\_\*(X),G)$. For any $A$ and $B$, we can then describe the dual $\operatorname{Hom}([A,B],G)$ as $[F(A,B),I\_G]$.


In the special case that either $A$ or $B$ is a finite spectrum, however, we can do better than this. If $DA=F(A,S)$ is the Spanier-Whitehead dual of $A$, we have $$[F(A,B),I\_G]=[DA\wedge B,I\_G]=[DA,F(B,I\_G)].$$


Since $F(B,I\_G)$ is just what we call $I\_GB$, we can thus describe the dual of $[A,B]$ as $[DA,I\_GB]$.


