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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

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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 $D\_\infty$.
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