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?

  • 1
    $\begingroup$ That's because they are defined stably, i.e. using an appropriate colimit. You have to look at what happens when you stabilize. $\endgroup$ – Michal Adamaszek Jun 29 '18 at 8:59
  • $\begingroup$ Do you know a good introduction to this theory? Could I find it in Hatcher's book? $\endgroup$ – Pepe Jun 29 '18 at 9:16
  • 1
    $\begingroup$ Proposition 4F.1 is exactly what you need for the case of suspension spectra, but the general argument for CW spectra is similar. $\endgroup$ – Michal Adamaszek Jun 29 '18 at 9:25
  • $\begingroup$ Note that in spectra a finite wedge sum is a finite product. $\endgroup$ – Qiaochu Yuan Jun 29 '18 at 15:24
  • $\begingroup$ 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. $\endgroup$ – Pepe Jun 29 '18 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.