MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Define an $S^{1}$-spectrum $E$ to be a sequence of pointed simplicial sets $E_{n},\\ n=0,1,2...$ with assembly morphisms $\sigma_{n}:S^{1}\wedge E_{n}\rightarrow E_{n+1}$. An $S^{1}$-spectrum $E$ is now called $\textit{fibrant}$ if all the simplicial sets $E_{n}$ are Kan-fibrant and the adjoint $E_{n}\rightarrow\Omega (E_{n+1})$ of $\sigma_{n}$ is a simplicial weak equivalence. My questions is now if the coproduct $\vee {E_{i}}$ of fibrant spectra $E_{i}$ is fibrant again?

Thank you

share|cite|improve this question

No. Here are two reasons:

  • The coproduct of two pointed Kan complexes is usually not a Kan complex (say, if the spaces are connected): we can map $\Lambda^2_1$ into "both summands" and the map will not extend to $\Delta^2$.
  • The functor $\Omega$ will basically never commute with wedge sums (even up to homotopy). For example, there is a $\Omega$-spectrum $E = H\mathbb{F}\_2$ with $E\_0 = \mathbb{F}\_2$ and $E\_1 = K(\mathbb{F}\_2, 1)$; so $\pi\_0(E\_0 \vee E\_0)$ is a three-element set but $\pi\_0(\Omega(E\_1 \vee E\_1))$ is the coproduct of $\mathbb{F}\_2$ with itself in the category of groups, which is the infinite dihedral group $D\_\infty$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.