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](https://mathoverflow.net/questions/117684/are-spectra-really-the-same-as-cohomology-theories) 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.



unwinding the definition of $H_i(KU)$ as a map of spectra $\mathbb{S}^i \to HZ \wedge KU$
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.


