# Invertibility of suspension in spectra

I know that spectra are supposed to be designed so that suspension is invertible up to homotopy, but I'm having trouble articulating exactly why this is the case.

If $E$ is a spectrum and $\Sigma E$ is the spectrum where $(\Sigma E)_n = E_{n+1}$, then there is a map $E \to \Sigma E$ where $E_n \to \Sigma E_n \to E_{n+1} = (\Sigma E)_n$ is defined using the structure map. I can think of two ways to proceed.

Option 1: Find a map $\Sigma E \to E$, i.e. a map $E_{n+1}\to E_n$ (or some suspension of this). I don't really know how to get a map that decreases degree, though; the adjoint of $\Sigma E_n \to E_{n+1}$ is $E_n \to \Omega E_{n+1}$ which doesn't help.

Option 2: Show that $E \to \Sigma E$ is a weak equivalence by showing that $\pi_r E \to \pi_r \Sigma E$ is an isomorphism. This actually seems wrong: if $E$ is an $\Omega$-spectrum then $\pi_{n+r}E_n = \pi_{n+1+r}E_{n+1}$ for large $n$, so $$\pi_r E = \lim_n \pi_{n+r}E_n = \lim_n \pi_{n+r+1}E_{n+1} = \lim_n \pi_{n+r+1}(\Sigma E)_n = \pi_{r+1}(\Sigma E).$$

• I'm a little confused. You seem to be trying to show that the maps $E_n \rightarrow E_n \wedge S^1 \rightarrow E_{n+1}$ define an equivalence of spectra. This would be saying that suspension is equivalent to the identity, which isn't what you want. There are two ways to define suspension: One is kind of silly, and you just shift the spaces all up by one. This is obviously invertible, since you can just shift the spaces back down. However, you could also define suspension by smashing with $S^1$. The trick is to show these two are equivalent. – Zach L. Nov 10 '14 at 4:00
• OK, I guess then I'm confused why "shift back down" is an actual map of spectra -- you'd need maps $(\Sigma E)_n = E_{n+1} \to E_n$ (or some suspension of this), right? – xenakis Nov 10 '14 at 4:21
• "Shift down" is a functor from the category of spectra to itself. It does not induce a map on every individual spectrum, but sends each spectrum to another spectrum (namely its shift). – Qiaochu Yuan Nov 10 '14 at 7:04
• Oh! Makes sense now, thanks. – xenakis Nov 10 '14 at 8:01
• @xenakis Are you from Chania, Greece? – user175343 Feb 23 '15 at 19:34