I am reading Hatcher's treatment of the Adam's spectral sequence. <http://pi.math.cornell.edu/~hatcher/SSAT/SSch2.pdf>


On page 20, he states "Thus for each $i$ the groups $\pi\_i(Z^k)$ are zero for all sufficiently large $k$. The same is true for the groups $\pi^Y\_i (Z^k)$ when $Y$ is a finite spectrum, since a map
$\Sigma^i Y \to Z^k$ can be homotoped to a constant map one cell at a time if all the groups $\pi\_j (Z^k)$ vanish for $j$ less than or equal to the largest dimension of the cells of $\Sigma^i Y$.''


Context: $Z^k$ is a CW spectrum of finite type. Also, $\pi^Y\_n(Z) = [ \Sigma^n Y, Z ] = colim\_k [ \Sigma^{n} Y\_k, Z\_k]$, where the $Y\_k$ and $Z\_k$ are spaces. 


I don't really understand the details of his statement. So my questions are 


1) Can someone explain the details of this more?


2) (I'd prefer) Can we say instead that this holds because we can write $Y$ as a finite limit of sphere spectra, and finite limits and filtered colimits commute? And if so, how does one write $Y$ as a finite limit of sphere spectra?... Does this work exactly the same as if we had spaces?



moving from sphere spectrum to finite spectrum
Let $\{Y,Z\}$ denote the homotopy group of maps between a spectrum $Y$ and a spectrum $Z$. We are assuming that one has a sequence of spectra $Z^1, Z^2, \dots$ such that for any fixed $i$, $\{S^i,Z^k\} = 0$ for $k>>0$, and we want to show that, if $Y$ is a finite spectrum, then $\{Y,Z^k\} = 0$ for $k>>0$. 


An easy way to see this is by induction on the number of cells of $Y$, with inductive step as follows: if $Y$ is obtained from $X$ by attaching an $n$--cell, then, by basic homotopy theory, the cofibration sequence $X \rightarrow Y \rightarrow S^n$ induces $\{S^n,Z^k\} \rightarrow \{Y,Z^k\} \rightarrow \{X,Z^k\}$ exact in the middle. By assumption and inductive hypothesis, the first and last of these are 0 for $k>>0$, and thus the so is the middle term. [This is clearer than Hatcher, who is sort of saying the same thing much more awkwardly.]


