I asked this on mathstackexchange but didn't get any response (or many views) so I'm asking it here, although clearly it belongs over there.

In the answer to this question on mathoverflow, it says: "The integral homology group $H_i(KU)$ is 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 $E=\Sigma H\mathbb{Z}$?) 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. I'm still learning how to best think of maps between spectra.

  • $\begingroup$ On one hand, homology can actually be define as such a direct limit. On the other hand as you say, it can be defined as homotopy of $E \wedge X$. I believe it is in an old paper of Whitehead that he proved they are equal. This equivalence is shown but not proven in Adams blue book. $\endgroup$ Oct 9, 2015 at 15:18

1 Answer 1


I'll define a naive prespectrum to be a system of spaces $X_i$ with maps $\Sigma X_i\to X_{i+1}$. That seems to be what you want to work with, but it is not technically very satisfactory. However, if $\mathcal{C}$ is one of the fancier categories of spectra, then one can define $FX$ to be the homotopy colimit in $\mathcal{C}$ of the objects $\Sigma^{-i}\Sigma^\infty X_i$, and that will give an object of $\mathcal{C}$ that has the homotopy type that you want. From this perspective, the real point is that $FX \wedge FY$ is the homotopy colimit of the diagram of objects $\Sigma^{-i-j}\Sigma^\infty X_i\wedge Y_j$, and that colimit can be broken down in three different ways:

  • Take the colimit over $i$ first, then the colimit over $j$
  • Take the colimit over $j$ first, then the colimit over $i$
  • Take the colimit of the diagonal terms $\Sigma^{-2i}\Sigma^\infty X_i\wedge Y_i$.

It is a fairly formal argument to show that these are all the same, and so $$ X_nY = \text{colim}_{i,j}\pi_{n+i+j}(X_i\wedge Y_j) = \text{colim}_i \pi_{n+2i}(X_i\wedge Y_i) = \text{colim}_i X_{n+i}Y_i = \text{colim}_i Y_{n+i}X_i. $$

If you want to work exclusively with naive prespectra, you can use the "handicrafted smash product" $$ (X\wedge Y)_{2i} = X_i\wedge Y_i $$ $$ (X\wedge Y)_{2i+1} = X_i\wedge Y_{i+1} $$ It is unpleasant to prove any properties based on this definition, but it is adequate for the calculation that you asked about.

  • $\begingroup$ Thank you very much for clarifying. I guess I should change what I first think about when I see $X_nY$, and instead of $[\Sigma^n \mathbb{S}, X \wedge Y]$, I should think $colim_i X_{n+i}Y_i$ $\endgroup$
    – usr0192
    Oct 9, 2015 at 16:08

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