Loop-Suspension adjunction unit is stable equivalence

algebraic-topologyhigher-homotopy-groupshomotopy-theoryspectrastable-homotopy-theory

Let $X$ be a sequential spectrum, then the unit of the $(\Sigma,\Omega)$-adjunction yields a map $\eta : X \to \Omega \Sigma X$. The authors of the book Foundations of Stable Homotopy Theory (p. 48) then claim that $\eta$ is a $\pi_*$-isomorphism because $\pi_*(\eta)$ is part of the colimit sequence defining the homotopy groups of spectra:
$$ \pi_{n+a}(X_a) \to \pi_{n+a}(\Omega \Sigma X_a) \cong \pi_{1+n+a}(\Sigma X_a) \to \pi_{1+n+a}(X_{1+a}).$$ Why does this observation show that $\eta$ is a $\pi_*$-isomorphism? Does this already tell us that the colimit sequence is somehow compatible with the one for $\pi_n(\Omega \Sigma X)$?

(I know a different argument using that suspension and loop space are $\pi_*$-equivalences along with triangle identities but I want to understand this more direct argument.)

Best Answer

This is a general fact about sequential colimits (or even more generally colimits over posets). $\newcommand\colim{\operatorname{colim}}$

Let $A : \mathbb{N}\to \mathcal{C}$ be some sequential diagram. $\mathbb{N}$ has an endomorphism $s$ defined by $s(n)=n+1$. There is a corresponding natural transformation $\sigma : A \to s^*A$, which at $n$ is the map $f_n : A_n\to A_{n+1}$ which is $A(n\le n+1)$, the transition map in the sequential diagram.

In fact, more generally, let $t : \mathbb{N}\to \mathbb{N}$ be any nondecreasing map, then $t$ gives an endomorphism of $\mathbb{N}$ as a category, so we can consider the pullback $t^*A$. If $t$ additionally satisfies $t(n)\ge n$ for all $n$, then there is a natural transformation $\tau: A\to t^*A$ which at $n$ is the unique map $A_n\to A_{t(n)}$ in the diagram $A$.

Now if we assume that $t$ is cofinal (for all $n\in \mathbb{N}$, there exists $N\in\mathbb{N}$ such that $t(N)\ge n$), the following is true.

Proposition. $\colim A$ exists if and only if $t^*\colim A$ exists. In this case if $t(n)\ge n$ for all $n$, so $\tau$ exists, then $\tau$ induces a morphism $\tilde{\tau}:\colim A\to \colim t^* A$ on passing to colimits which is an isomorphism.

Proof.

It suffices to give a natural isomorphism between $\newcommand\cocone{\operatorname{co-cone}}\cocone_A$ and $\cocone_{t^*A}$, the functors that the colimits represent. Given some cocone to $A$, $(X, i_n: A_n\to X)$, we get a cocone to $t^*A$ by taking $(X, i_{t(n)}: A_{t(n)}\to X)$. However, it turns out that we can also go backwards, since $t$ is cofinal. If we are given $(Y,j_m : A_{t(m)}\to Y)$, we can define $i_n : A_n\to Y$ by taking $N$ large enough that $t(N)\ge n$, and then defining $i_n$ to be the composite $A_n \to A_{t(N)} \xrightarrow{j_N} Y$. This is well defined regardless of the choice of $N$ because $\mathbb{N}$ is a poset and $(Y,j_m)$ are a cocone to $t^* A$. This gives us a cocone $(Y, i_n)$ to $A$. These operations are inverse to each other, so we have a natural isomorphism as desired.

Finally we need to show that when $t(n)\ge n$ for all $n$, and if $\colim A$ exists then $\tau$ induces an isomorphism on the colimits. First let's fix some notation. Let $i_n : A_n\to \colim A$ be the inclusions for the colimit of $A$, let $j_n : A_{t(n)}\to \colim t^*A$ be the inclusions of the colimit of $t^*A$, and let $\tau_n: A_n\to A_{t(n)}$ be the components of $\tau$. Then we observe that $\tilde{\tau}$ is defined by applying the universal property of the colimit $\colim A$ to the cocone $(\colim t^*A, j_n \circ \tau_n)$. But we know (up to isomorphism) what the $j_n$ are. By the first half of this proof the $j_n$ are $i_{t(n)}$, so $$j_n\circ \tau_n = i_{t(n)}\circ \tau_n = i_n,$$ since $\tau_n : A_n\to A_{t(n)}$ is a map in the diagram $A$ and $i$ is a cocone to $A$. Thus after potentially composing with an isomorphism $(\colim t^*A, j_n)\to (\colim A, i_{t(n)})$, $\tilde{\tau}$ is actually given by the colimiting cone itself, so we have that for some isomorphism $\phi$, $\phi\circ \tilde{\tau} = 1_{\colim A}$, and thus $\tilde{\tau}$ is an isomorphism itself (namely $\phi^{-1}$). $\blacksquare$

Edited to fix the final argument

Build a modified version of the stable homotopy group colimit sequence as follows.

Fix $a$, let $P_{2n} = \pi_{n+a}(X_a)$, let $P_{2n+1} = \pi_{n+a}(\Omega\Sigma X_a)$, let the maps $f_{2n} : P_{2n}\to P_{2n+1}$ be $\pi_{n+a}(\eta_{X_a})$, and let the maps $f_{2n+1} : P_{2n+1}\to P_{2n+2}$ be given by the composite of $\pi_{1+n+a}$ applied to the structure map $\Sigma X_a\to X_{a+1}$ with the isomorphism $\pi_{n+a}(\Omega\Sigma X_a)\cong \pi_{n+1+a}(\Sigma X_a)$.

Let $d: \mathbb{N}\to \mathbb{N}$ be given by $d(n) =2n$, and let $\delta : P\to d^*P$ be the corresponding natural transformation. Note that $d^*P$ is precisely the sequence that usually defines the stable homotopy groups.

Now consider the $\sigma$ natural transformation, $\sigma_P$, for this sequence. If we restrict to even $n$, then $\sigma_P$ corresponds to $\pi_{\bullet+a}(\eta)$. In other words $d^*\sigma_P = \pi_{\bullet + a}(\eta)$. We have a commutative diagram $$ \require{AMScd} \begin{CD} P @>\sigma_P>> s^* P\\ @V\delta VV @V\delta VV \\ d^*P @>>\pi_{\bullet+a}(\eta)> d^*s^*P.\\ \end{CD} $$ Termwise this commutative diagram is $$ \require{AMScd} \begin{CD} P_n @>>> P_{n+1}\\ @VVV @VVV \\ P_{2n}=\pi_{n+a}(X_a) @>>\pi_{n+a}(\eta_{X_a})> P_{2n+1}=\pi_{n+a}(\Omega \Sigma X_a).\\ \end{CD} $$

Now if we take colimits of this diagram of functors, three of the maps, $\sigma_P$ and both $\delta$s induce isomorphisms on the colimit. Therefore the last map does as well.

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