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In chapter 22.1 of May's A Concise Course in Algebraic Topology, he claims that the prespectrum $\{T_n\}$ of spaces where each $T_n$ is $(n-1)$-connected yields a reduced homology theory by setting $\tilde{E}_q(X) = \operatorname{colim}_n \pi_{q+n}(X \wedge T_n)$, where each map in the colimit is given by $$\pi_{q+n}(X \wedge T_n) \xrightarrow{\Sigma} \pi_{q+n+1}(\Sigma(X \wedge T_n)) \cong \pi_{q+n+1}(X \wedge \Sigma T_n) \xrightarrow{\operatorname{id} \wedge \sigma} \pi_{q+n+1}(X \wedge T_{n+1}),$$ where $\sigma: \Sigma T_n \to T_{n+1}$ are the maps included in the data of the prespectrum. I understand his proof of the exactness and additivity axioms, but I am confused about the suspension axiom.

He defines a suspension map by $$\pi_{q+n}(X \wedge T_n) \xrightarrow{\Sigma} \pi_{q+n+1}(\Sigma(X \wedge T_n)) \cong \pi_{q+n+1}((\Sigma X) \wedge T_n)$$ and claims that these maps commute with the maps in the colimit. However, if I'm not mistaken, doesn't the diagram $$\require{AMScd} \begin{CD} \pi_{q+n}(X \wedge T_n) @>{\Sigma}>> \pi_{q+n+1}((\Sigma X) \wedge T_n) \\ @VVV @VVV \\ \pi_{q+n+1}(X \wedge T_{n+1}) @>{\Sigma}>> \pi_{q+n+2}((\Sigma X) \wedge T_{n+1}) \end{CD}$$ anticommute, since the map $\Sigma \Sigma S^{q+n} \to (\Sigma X) \wedge T_{n+1}$ obtained by chasing the diagrams through both directions have the last two suspension coordinates permuted, thus changing the sign by $-1$? It's not a big deal for the purpose of proving this axiom, since we can just alternate $\Sigma$ with $-\Sigma$ depending on the parity of $q+n$ to get a perfectly good suspension homomorphism which is natural in $X$, but I just wanted to make sure I wasn't missing anything here.

e_s_
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1 Answers1

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You're right that this square does not commute, only up to sign. In fact, in stable homotopy theory one must make a clever choice which suspensions one uses at which moment. Let me illustrate this. May defines $\Sigma X:= X\wedge S^1$ and lets in a prespectrum $T$ the structure maps be given by maps of the form $T_n\wedge S^1\to T_{n+1}$. This in itself already gives you a slight inconvenience when defining the suspension $\Sigma T$ of $T$. Namely, we want $(\Sigma T)_n := \Sigma(T_n):= T_n\wedge S^1$, but now the structure maps have to be given by $$(\Sigma T)_n\wedge S^1\cong T_n\wedge S^1\wedge S^1\xrightarrow{\mathrm{switch}}T_n\wedge S^1\wedge S^1\to T_{n+1}\wedge S^1=:(\Sigma T)_{n+1},$$ where the switch map $S^1\wedge S^1\to S^1\wedge S^1, s\wedge t\mapsto t\wedge s$ switches the coordinates. We need to add this switch map and remember it always, just because we let our suspension act from the same side as the spheres in the structure maps of a prespectrum. This is however also the reason your square does not commute, and therefore I prefer the following set-up (which is what e.g. Stefan Schwede uses).

We let in a prespectrum $T$ the spheres act from the left side, so the structure maps are of the form $S^1\wedge T_n\to T_{1+n}$. (We write $1+n$ instead of $n+1$, to remember the side on which ''the extra coordinate is added''. In stable homotopy theory, we joke that addition is not commutative.) We let the suspension of a space $X$ be $\Sigma X:=X\wedge S^1$. Now we can define $E_q(X)=\mathrm{colim}_n\, \pi_{n+q}(T_n\wedge X)$, with the colimit system having structure maps $$ \pi_{n+q}(T_n\wedge X)\xrightarrow{S^1\wedge-}\pi_{1+n+q}(S^1\wedge T_n\wedge X)\to\pi_{1+n+q}(T_{1+n}\wedge X). $$ We define the suspension isomorphism to be induced by the composite $$ \pi_{n+q}(T_n\wedge X)\xrightarrow{-\wedge S^1}\pi_{n+q+1}(T_n\wedge (X\wedge S^1))=\pi_{n+q+1}(T_n\wedge \Sigma X). $$ Now the analogue of the square in your question does actually commute on the nose, and in many more situations, this convention of letting spheres act on the left in prespectra and suspensions (of spaces or spectra) being defined by smashing on the right makes sure that spheres that serve different purposes do not get intermingled and do not require extra switch isomorphisms (and their signs) to untangle them.

Daniël Apol
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