Let $E$ be a multiplicative spectrum, connective, and assume $\pi_0E$ is cyclic.
I want to prove that $$H^0(E;\pi_0E)\cong \hom_{\pi_0E}(H_0(E;\pi_0E);\pi_0E)$$
Recall that $H^0(E;\pi_0E):= [E; K(\pi_0E)]$ and $H_0(E;\pi_0E):=\pi_0(E \wedge K(\pi_0E))$.
My attempt: It's equivalent to prove that the pairing $$\langle H^0(E;\pi_0E) , H_0(E;\pi_0E) \rangle \to \pi_0E$$ is perfect. But I don't know how to proceed with computation since I don't know much about the cohomology $H^0(E;\pi_0E)$, nor the maps $[E; K(\pi_0E)]$.
Can someone give me some hints?
By the Hurewicz theorem for spectra, if $X$ is a connective spectrum, then $H_n(X;\mathbb{Z})=0$ for $n<0$ and $H_0(X;\mathbb{Z})\cong \pi_0(X)$. It follows by the universal coefficient theorems that for any abelian group $A$, $H_0(X;A)\cong \pi_0(X)\otimes A$ and $H^0(X;A)\cong \operatorname{Hom}(\pi_0(X),A)$.
Applying this with $X=E$ and $A=\pi_0(E)$, it is then trivial to see that if $\pi_0(E)$ is cyclic, the two groups you are comparing are both isomorphic to $\pi_0(E)$.