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In the last chapter of his Concise course in algebraic topology, May states (without proof or reference) that for an arbitrary collection $(X_i)_{i\in I}$ of spectra the following hold:

$\pi_n(\prod_{i\in I} X_i)=\prod_{i\in I} \pi_n(X_i)$

$\pi_n(\bigvee_{i\in I} X_i)=\sum_{i\in I} \pi_n(X_i)$

The first one is clear, since the same holds for spaces. But the homotopy of a wedge sum can be quite complicated in general, doesn't it? Is this an exclusive property of spectra? Why does it hold?

Ken
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Pepe
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    That's because they are defined stably, i.e. using an appropriate colimit. You have to look at what happens when you stabilize. – Michal Adamaszek Jun 29 '18 at 08:59
  • Do you know a good introduction to this theory? Could I find it in Hatcher's book? – Pepe Jun 29 '18 at 09:16
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    Proposition 4F.1 is exactly what you need for the case of suspension spectra, but the general argument for CW spectra is similar. – Michal Adamaszek Jun 29 '18 at 09:25
  • Note that in spectra a finite wedge sum is a finite product. – Qiaochu Yuan Jun 29 '18 at 15:24
  • This is exactly what I'm trying to comprehend. I'm searching for an explanation on space-level why this holds. But it gets a bit nasty with all the limits and colimits. – Pepe Jun 29 '18 at 15:42
  • A textbook account can be found in Switzer's book Algebraic Topology - Homotopy and Homology, Corollary 8.36. – Ken Aug 09 '22 at 02:39

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