2 added 517 characters in body

I claim that, in the equivalences you stated, duality has nothing to do with it. Specifically, if your viewpoint were correct, then for finite $G$, $BG(1)$ would be non-canonically isomorphic to $K(G,1)$. In fact, I claim that they are canonically isomorphic. Furthermore, the $\mathbb Z$ in $\mathbb CP^{\infty}$ is canonically isomporphic to the fundamental group, not the character group, of $U(1)$.

Argument: Since both these spaces represent functors, it suffices to consider the underlying functors. Eilenberg Mac-Lane spaces correspond to cohomology functors. It is easy to prove using Cech cohomology that cohomology with coefficients in $G$ naturally classifies principal $G$-bundles. This, of course, is exactly what the classifying space classifies - not dual to what the classifying space classifies.

$\mathbb CP^{\infty}$: There is an exact sequence $0\to\mathbb Z \to \mathbb C^+ \to \mathbb C^\times\to 0$, giving a map $H^1(X,\mathbb C^\times)\to H^2(X,\mathbb Z)$. The image is discrete while the kernel, a quotient of $H^1(X,\mathbb C^{+})$, is connected, so the map is exactly the quotient by the connected component of the identity.

$H^1(X,\mathbb C^\times)$ classifies principal $\mathbb C^\times$ spaces. Continuously moving the bundle around in it corresponds to continuously deforming the bundle. These bundles up to derivation are exactly what $BU(1)$ classifies.

EDIT: Idea/sketch for a general proof of this equivalence: Let $G_n$ be the group of principal $G$-bundles on $S^{n}$. Then for some reason this should be equivalent to $\pi_{n-1}(G)$. Now, the values everywhere of good functors on the category of CW complexes (specifically, representable ones) depend only on the values they take on spheres. So suppose a group had only one nontrivial $G_n$. The principal $G$-spaces functor would then be equivalent to $H^n(X,G_n)$, giving an equivalence of classifying spaces.

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I claim that, in the equivalences you stated, duality has nothing to do with it. Specifically, if your viewpoint were correct, then for finite $G$, $BG(1)$ would be non-canonically isomorphic to $K(G,1)$. In fact, I claim that they are canonically isomorphic. Furthermore, the $\mathbb Z$ in $\mathbb CP^{\infty}$ is canonically isomporphic to the fundamental group, not the character group, of $U(1)$.

Argument: Since both these spaces represent functors, it suffices to consider the underlying functors. Eilenberg Mac-Lane spaces correspond to cohomology functors. It is easy to prove using Cech cohomology that cohomology with coefficients in $G$ naturally classifies principal $G$-bundles. This, of course, is exactly what the classifying space classifies - not dual to what the classifying space classifies.

$\mathbb CP^{\infty}$: There is an exact sequence $0\to\mathbb Z \to \mathbb C^+ \to \mathbb C^\times\to 0$, giving a map $H^1(X,\mathbb C^\times)\to H^2(X,\mathbb Z)$. The image is discrete while the kernel, a quotient of $H^1(X,\mathbb C^{+})$, is connected, so the map is exactly the quotient by the connected component of the identity.

$H^1(X,\mathbb C^\times)$ classifies principal $\mathbb C^\times$ spaces. Continuously moving the bundle around in it corresponds to continuously deforming the bundle. These bundles up to derivation are exactly what $BU(1)$ classifies.