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?


2 Answers 2


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.]

  • $\begingroup$ So basically, exactly the same as if everything were spaces, except that since we are in the stable homotopy category, fibrations and cofibrations are the same so that we can get the exact sequence? $\endgroup$
    – Elise
    Jan 17, 2020 at 15:36
  • $\begingroup$ @Elise Yes. But even better: the cofibration is in the first variable, so it is exact in the world of spaces too. No fibrations needed here. $\endgroup$ Jan 17, 2020 at 16:13

For convenience I will set $Z = Z^k$, where $Z^k$ is as in your notation.

Case 1: $Y$ is the suspension spectrum of a based finite complex $U$ having dimension $n$. Then $\pi^Y_\ast(Z)$ is given by the homotopy groups of the function space of based maps $F(U,\Omega^\infty Z)$, where $\Omega^\infty Z$ is the infinite loop space associated with $Z$.

Filter $U$ by its skeleta $U^{(j)} \subset U$. Then we have a sequence of fibrations $$ F(U,\Omega^\infty Z) = F(U^{(n)},\Omega^\infty Z) \to F(U^{(n-1)},\Omega^\infty Z)\to \cdots \to F(U^{(0)},\Omega^\infty Z)\, . $$ The fibers of the map at stage $j$ in the sequence are identified with $F(U^{(j)}/U^{(j-1)},\Omega^\infty Z)$ and the latter is a finite product of copies of $F(S^j,\Omega^\infty Z)$.

The homotopy groups of $F(S^j,\Omega^\infty Z)$ are just the homotopy groups of $Z$ shifted by $j$. So, by your assumptions, for any index $i$ there is a $k$ with $Z= Z^k$ such that $\pi_\ell$ of $F(S^j,\Omega^\infty Z)$ vanishes for $\ell\le i$. Consequently, the fiber of the $j$th fibration is $i$-connected. Furthermore, a similar argument shows that $F(U^{(0)},\Omega^\infty Z)$ is $i$-connected. It follows that $F(U,\Omega^\infty Z)$ is also $i$-connected.

Case 2: $Y$ is an arbitrary finite spectrum. Then there is a non-negative integer $t$ such that $\Sigma^t Y$ is the suspension spectrum of a finite complex $U$. So, $F(Y,Z) = F(U,\Sigma^t Z)$. The argument then works as above with $Z$ replaced by $\Sigma^t Z$.

Remark: What the above in effect shows is this: let $Y$ be a finite dimensional cell spectrum of dimension $s\in \Bbb Z$ and let $Z$ be an $r$-conected spectrum. Then the function spectrum $F(Y,Z)$ is $(r-s)$-connected.


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