In the answer to this question on mathoverflow, it says The integral homology group $H_i(KU)$, the direct limit of

$$\dots \to H_{2n+i}(BU)\to H_{2n+2+i}(BU)\to\dots,$$

My question is why? Here is my understanding: Given two spectra $E, X$, I believe the definition of the homology group $E_i(X)$ is $$E_i(X):=[\Sigma^i \mathbb{S}, E \wedge X]_{stable}=colim_j[S^{i+j}, (E \wedge X)_j]$$ where the square brackets denote homotopy classes of maps. In the case at hand, $E=H\mathbb{Z}$ (or maybe $E=H\mathbb{Z}$?) is the spectrum whose $n$th space is the Eilenberg-MacLane space $K(\mathbb{Z}, n)$, and $X=KU$ is the spectrum whose $n$ space is either $BU \times \mathbb{Z}$ or $BU$, depending on if $n$ is even or odd, respectively. I have no idea what the $n$th space of $HZ \wedge KU$ is, since I am confused by smash product of spectra.

  • $\begingroup$ A full derivation of $H_i(KU)$ from the homology of $BU$ can be found in Switzer, Theorem 16.25. $\endgroup$ – Bruno Stonek Nov 26 '18 at 15:27

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