# cohomology is dual to homology of a spectrum if homology is free

Let $E$ be a multiplicative spectrum (and $X$ a space with $H_n(X; \mathbb{Z})$ free abelian for every $n$). The following excerpt is taken from the notes here claim that item (1) below easily implies (2). Can someone tell me how? I suspect it is some thing involving the adjunction $[Z, E^X]=[Z \otimes X, E]$, and/or the fact that the homotopy groups of $E \otimes \Sigma X_+$ are $E_*(X)$ and the homotopy groups of $E^X$ are the cohomology groups $E^*(X)$. Perhaps part 1 is saying homology groups of $E$ are free, and so then by universal coefficient theorem (does this hold for spectra?), the cohomology groups are dual to the homology. But I don't know if this is right, or if there is a way to state this idea precisely.

• I think the point is that $E^X$ is also the spectrum of morphisms $E \otimes \Sigma^{\infty} X_{+} \to E$ of $E$-modules, but I could be wrong. – Qiaochu Yuan Jul 31 '14 at 6:58
• i was reading this last week and Qiaochu's argument is what i ended up with as well – Arpon Aug 2 '14 at 15:01

Based on Qiaochu's comment, here is my reasoning: We want to show $E^X$ and $\prod \Sigma^{-n} E$ have the same homotopy groups (that's what equivalence means). Since these are $E$-modules, a map of spectra $S \to E^X$ corresponds to a map of $E$-modules $E \to E^X$.
and a map of $E$-modules $E \to \prod \Sigma^{-n} E$ is the same thing as a map of spectra $S \to \prod \Sigma^{-n} E$, and so we see that $E^X$ and $\prod \Sigma^{-n} E$ have the same homotopy groups.